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| author | Trupti Kini | 2016-04-21 23:30:25 +0600 |
|---|---|---|
| committer | Trupti Kini | 2016-04-21 23:30:25 +0600 |
| commit | 74fc7263ac6ab1f65240e43d8a9635aae379e849 (patch) | |
| tree | 3f3b98c932a72c1c8103317be141f1ce62b45dbc | |
| parent | 92aa8e6c8882e1b9cba68371aec84017181e2cce (diff) | |
| download | Python-Textbook-Companions-74fc7263ac6ab1f65240e43d8a9635aae379e849.tar.gz Python-Textbook-Companions-74fc7263ac6ab1f65240e43d8a9635aae379e849.tar.bz2 Python-Textbook-Companions-74fc7263ac6ab1f65240e43d8a9635aae379e849.zip | |
Added(A)/Deleted(D) following books
A College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch15_1.ipynb
A College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch16_1.ipynb
A College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch17_1.ipynb
A College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch18_1.ipynb
A College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch19_1.ipynb
A College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch20_1.ipynb
A College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch21_1.ipynb
A College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch22_1.ipynb
A College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch23_1.ipynb
A College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch24_1.ipynb
A College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch25_1.ipynb
A College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch26_1.ipynb
A College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch27_1.ipynb
A College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch28_1.ipynb
A College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch29_1.ipynb
A College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch30_1.ipynb
A College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/screenshots/1RefractionOfLaserLight_1.png
A College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/screenshots/2PropertiesOfImage_1.png
A College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/screenshots/3PositionOf1DarkFringe_1.png
A principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/Chapter10_1.ipynb
A principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/Chapter28_1.ipynb
A principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/Chapter29_1.ipynb
A principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter11_1.ipynb
A principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter12_1.ipynb
A principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter13_1.ipynb
A principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter14_1.ipynb
A principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter15_1.ipynb
A principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter16_1.ipynb
A principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter17_1.ipynb
A principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter18_1.ipynb
A principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter19_1.ipynb
A principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter1_1.ipynb
A principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter20_1.ipynb
A principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter21_1.ipynb
A principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter22_1.ipynb
A principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter23_1.ipynb
A principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter24_1.ipynb
A principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter25_1.ipynb
A principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter26_1.ipynb
A principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter27_1.ipynb
A principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter2_1.ipynb
A principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter3_1.ipynb
A principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter4_1.ipynb
A principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter5_1.ipynb
A principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter6_1.ipynb
A principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter7_1.ipynb
A principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter8_1.ipynb
A principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter9_1.ipynb
A principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/screenshots/image11.png
A principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/screenshots/image12.png
A principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/screenshots/image13.png
_by_R._A._Serway_and_J._S._Faughn/screenshots/1RefractionOfLaserLight_1.png?id=74fc7263ac6ab1f65240e43d8a9635aae379e849){kind=link}
_by_R._A._Serway_and_J._S._Faughn/screenshots/2PropertiesOfImage_1.png?id=74fc7263ac6ab1f65240e43d8a9635aae379e849){kind=link}
_by_R._A._Serway_and_J._S._Faughn/screenshots/3PositionOf1DarkFringe_1.png?id=74fc7263ac6ab1f65240e43d8a9635aae379e849){kind=link}
51 files changed, 28163 insertions, 0 deletions
diff --git a/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch15_1.ipynb b/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch15_1.ipynb new file mode 100644 index 00000000..1c77fdc3 --- /dev/null +++ b/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch15_1.ipynb @@ -0,0 +1,179 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 15 : Electric forces and electric fields" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 15.1 Page No : 502" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The attractive force = 8.19e-08 N\n", + "The gravitational force = 3.61e-47 N\n" + ] + } + ], + "source": [ + "k_e=8.99*10**9\n", + "e=1.6*10**-19\n", + "r=5.3*10**-11\n", + "F_e= (k_e*e*e)/(r*r)\n", + "print \"The attractive force = %0.2e N\"%F_e\n", + "G=6.67*10**-11\n", + "m_e=9.11*10**-31\n", + "m_p=1.67*10**-27\n", + "F_g=(G*m_e*m_p)/(r*r)\n", + "print \"The gravitational force = %0.2e N\"%F_g" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 15.2 Page No : 503" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The force = 5.62e-09 N\n", + "The force = 1.08e-08 N\n" + ] + } + ], + "source": [ + "k_e=8.99*10**9 #N.m**2/c**2\n", + "q2=2*10**-9# = %0.2f c\n", + "q3=5*10**-9# = %0.2f c\n", + "r1=4#in m\n", + "F_23=(q2*q3*k_e)/(r1*r1)\n", + "print \"The force = %0.2e N\"%F_23\n", + "q1=6*10**-9\n", + "r2=5#in m\n", + "F_13=(q1*q3*k_e)/(r2*r2)\n", + "print \"The force = %0.2e N\"%F_13" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 15.4 Page No: 507" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The magnitude of force = 3.20e-15 N\n" + ] + } + ], + "source": [ + "q=1.6*10**-19#in c\n", + "E=2*10**4# = %0.2f N/C\n", + "F=q*E\n", + "print \"The magnitude of force = %0.2e N\"%F" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 15.5 Page No: 509" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Magnitude of E1 = 3.93e+05 N/C\n", + "Magnitude of E2 = 1.80e+05 N/C\n", + "Magnitude in x direction = 1.80e+05 N/C\n", + "Magnitude in y direction = 2.49e+05 N/C\n", + "Angle = 54.17 degree\n" + ] + } + ], + "source": [ + "from math import degrees, atan\n", + "k_e=8.99*10**9 #N.m**2/c**2\n", + "q1=7*10**-6# = %0.2f C\n", + "q2=5*10**-6#in C\n", + "r1=0.4\n", + "r2=0.5\n", + "E1=(k_e*q1)/(r1**2)\n", + "E2=(k_e*q2)/(r2**2)\n", + "Ex=(k_e*q2)/(r2**2)\n", + "print \"Magnitude of E1 = %0.2e N/C\"%E1\n", + "print \"Magnitude of E2 = %0.2e N/C\"%E2\n", + "print \"Magnitude in x direction = %0.2e N/C\"%Ex\n", + "Ey=(3.93*10**5)+(-1.44*10**5)\n", + "print \"Magnitude in y direction = %0.2e N/C\"%Ey\n", + "phi=degrees(atan(Ey/Ex))\n", + "print \"Angle = %0.2f degree\"%phi\n", + "#Answer given in the book is wrong" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch16_1.ipynb b/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch16_1.ipynb new file mode 100644 index 00000000..0e794d3d --- /dev/null +++ b/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch16_1.ipynb @@ -0,0 +1,365 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 16 : Electrical Energy & Capacitance" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 16.1 Page No : 533" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The value of E = 4000.00 v/m\n" + ] + } + ], + "source": [ + "v_bminusv_a=-12\n", + "d=0.3*10**-2#in m\n", + "E=-(v_bminusv_a)/d\n", + "print \"The value of E = %0.2f v/m\"%E" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 16.2 Page No : 533" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "solution a\n", + "Electric potential from A to B = -40000.00 V\n", + "solution b\n", + "Change in electric potential = -0.00 joules\n", + "velocity = 2768514.16 m/s\n" + ] + } + ], + "source": [ + "from math import sqrt\n", + "print \"solution a\"\n", + "E=8*10**4#in V/m\n", + "d=0.5#in m\n", + "delta_V=-E*d\n", + "print \"Electric potential from A to B = %0.2f V\"%delta_V\n", + "print \"solution b\"\n", + "q=1.6*10**-19#in C\n", + "delta_PE=q*delta_V\n", + "print \"Change in electric potential = %0.2f joules\"%delta_PE\n", + "m_p=1.67*10**-27#in kg\n", + "vf=sqrt((2*-delta_PE)/m_p)\n", + "print \"velocity = %0.2f m/s\"%vf" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 16.3 Page No: 534" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Solution a\n", + "Magnitude of V1 = 112375.00 v\n", + "Magnitude of V2 = -35960.00 v\n", + "solution b\n", + "Magnitude of Vp = 76415.00 v\n", + "work done = 0.31 Joule\n" + ] + } + ], + "source": [ + "k_e=8.99*10**9 #N.m**2/c**2\n", + "q1=5*10**-6# in C\n", + "q2=-2*10**-6#in C\n", + "r1=0.4\n", + "r2=0.5\n", + "V1=(k_e*q1)/(r1)\n", + "V2=(k_e*q2)/(r2)\n", + "print \"Solution a\"\n", + "print \"Magnitude of V1 = %0.2f v\"%V1\n", + "print \"Magnitude of V2 = %0.2f v\"%V2\n", + "print \"solution b\"\n", + "vp=V1+V2\n", + "print \"Magnitude of Vp = %0.2f v\"%vp\n", + "q3=4*10**-6#in C\n", + "w=vp*q3\n", + "print \"work done = %0.2f Joule\"%w" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 16.4 Page No: 535" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Capacitance = 1.77e-12 farad\n" + ] + } + ], + "source": [ + "e0=8.85*10**-12#in c2/N.m2\n", + "A=2*10**-4#in m2\n", + "d=1*10**-3#in m\n", + "c=(e0*A)/d\n", + "print \"Capacitance = %0.2e farad\"%c" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 16.5 Page No : 535" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "capacitance = 4.50e-05 farad\n", + "voltage between battery = 2.16e-04 c\n" + ] + } + ], + "source": [ + "c1=3*10**-6\n", + "c2=6*10**-6\n", + "c3=12*10**-6\n", + "c4=24*10**-6\n", + "delta_v=18\n", + "c_eq=c1+c2+c3+c4\n", + "print \"capacitance = %0.2e farad\"%c_eq\n", + "q=delta_v*c3\n", + "print \"voltage between battery = %0.2e c\"%q" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 16.6 Page No : 536" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "solution a\n", + "capacitance = 1.60e-06 farad\n", + "solution b\n", + "voltage between battery = 2.88e-05 c\n" + ] + } + ], + "source": [ + "c1=3*10**-6\n", + "c2=6*10**-6\n", + "c3=12*10**-6\n", + "c4=24*10**-6\n", + "delta_v=18\n", + "print \"solution a\"\n", + "c_eq=1/((1/c1)+(1/c2)+(1/c3)+(1/c4))\n", + "print \"capacitance = %0.2e farad\"%c_eq\n", + "q=delta_v*c_eq\n", + "print \"solution b\"\n", + "print \"voltage between battery = %0.2e c\"%q" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 16.7 Page No: 536" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "solution a\n", + "capacitance = 2.00e-06 farad\n" + ] + } + ], + "source": [ + "c1=4*10**-6\n", + "c2=4*10**-6\n", + "print \"solution a\"\n", + "c_eq=1/((1/c1)+(1/c2))\n", + "print \"capacitance = %0.2e farad\"%c_eq" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 16.8 Page No: 537" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "solution a\n", + "Energy stored = 4671 volt\n", + "solution b\n", + "power = 240000 watt\n" + ] + } + ], + "source": [ + "from math import sqrt\n", + "Energy=1.2*10**3#in J\n", + "c=1.1*10**-4#in f\n", + "delta_v=sqrt((2*Energy)/c)\n", + "print \"solution a\"\n", + "print \"Energy stored = %0.f volt\"%delta_v\n", + "print \"solution b\"\n", + "Energy_deliverd=600#in j\n", + "delta_t=2.5*10**-3#in s\n", + "p=(Energy_deliverd)/delta_t\n", + "print \"power = %0.f watt\"%p" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 16.9 Page No: 538" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "solution a\n", + "Capacitance = 1.96e-11 farad\n", + "solution b\n", + "Voltage = 16000.0 volt\n", + "Maximum charge = 3.14e-07 columb\n" + ] + } + ], + "source": [ + "k=3.7\n", + "e0=8.85*10**-12#in c2/N.m2\n", + "A=6*10**-4#in m2\n", + "d=1*10**-3#in m\n", + "c=(k*e0*A)/d\n", + "print \"solution a\"\n", + "print \"Capacitance = %0.2e farad\"%c\n", + "print \"solution b\"\n", + "E_max=16*10**6#in v/m\n", + "delta_v_max=E_max*d\n", + "print \"Voltage = %0.1f volt\"%delta_v_max\n", + "Q_max=delta_v_max*c\n", + "print \"Maximum charge = %0.2e columb\"%Q_max" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch17_1.ipynb b/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch17_1.ipynb new file mode 100644 index 00000000..d0eee390 --- /dev/null +++ b/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch17_1.ipynb @@ -0,0 +1,311 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 17 : Current and resistance" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 17.1 Page No: 571" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "solution a : \n", + "Current = 0.83 Amp\n", + "solution b : \n", + "Number of electrons = 0.84 C\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "print \"solution a : \"\n", + "delta_q=1.67 # in c\n", + "delta_t=2 # in s\n", + "I=delta_q/delta_t\n", + "print \"Current = %0.2f Amp\"%I\n", + "print \"solution b : \"\n", + "N=5.22*10**18\n", + "N_q=(1.6*10**-19)*N\n", + "\n", + "print \"Number of electrons = %0.2f C\"%N_q" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 17.2 Page No: 573" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Solution a :\n", + "The drift speed = 2.46e-04 m/s=\n", + "Drift speed of electron = 1.15e+05 m/s\n" + ] + } + ], + "source": [ + "from math import sqrt\n", + "M=63.5 # IN G\n", + "rho=8.95\n", + "v=M/rho\n", + "electrons=6.02*10**23\n", + "n=(electrons*10**6)/v\n", + "I=10 # in c/s\n", + "q=1.60*10**-19 # in c\n", + "A=3*10**-6 # in m2\n", + "vd=(I)/(n*q*A)\n", + "print \"Solution a :\"\n", + "print \"The drift speed = %0.2e m/s=\"%vd\n", + "k_b=1.38*10**-23\n", + "T=293\n", + "m=9.11*10**-31\n", + "v_rms=sqrt((3*k_b*T)/m)\n", + "print \"Drift speed of electron = %0.2e m/s\"%v_rms" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 17.3 Page No: 578" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The resistance = 18.75 ohm\n" + ] + } + ], + "source": [ + "delta_v=120\n", + "I=6.4\n", + "R=(delta_v)/I\n", + "print \"The resistance = %0.2f ohm\"%R" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 17.4 Page No: 580" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Solution a : \n", + "Area = 3.24e-07 m**2\n", + "Resistance = 4.63 ohm/m\n", + "solution b : \n", + "The current = 2.16 Amps\n" + ] + } + ], + "source": [ + "from math import pi\n", + "r=0.321*10**-3\n", + "A=pi*(r*r)\n", + "print \"Solution a : \"\n", + "print \"Area = %0.2e m**2\"%A\n", + "rho=1.5*10**-6 # in ohm=m\n", + "l=rho/A\n", + "print\"Resistance = %0.2f ohm/m\"% l\n", + "print \"solution b : \"\n", + "Delta_v=10\n", + "I=(Delta_v)/l\n", + "print \"The current = %0.2f Amps\"%I\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 17.5 Page No: 582" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Temperature = 156.73 C\n" + ] + } + ], + "source": [ + "R=76.8\n", + "Ro=50\n", + "alpha=3.92*10**-3\n", + "t=(R-Ro)/(alpha*Ro)\n", + "T=t+20\n", + "print \"Temperature = %0.2f C\"%T" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 17.6 Page No: 583" + ] + }, + { + "cell_type": "code", + "execution_count": 16, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The current = 6.00 A\n", + "Power = 288.00 Watt\n" + ] + } + ], + "source": [ + "delta_v=50\n", + "R=8\n", + "I=(delta_v)/R\n", + "print \"The current = %0.2f A\"%I\n", + "P=I*I*R\n", + "print \"Power = %0.2f Watt\"%P" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 17.7 Page No: 585" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Number of bulbs = 32\n" + ] + } + ], + "source": [ + "I=20 # in A\n", + "delta_v=120\n", + "p_bulb=75 # inwatt\n", + "p_total=I*delta_v\n", + "N=p_total/p_bulb\n", + "print \"Number of bulbs = %d\"%N" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 17.8 Page No: 587" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Energy = 2.40 kwh\n", + "Cost = 0.29 dollars\n" + ] + } + ], + "source": [ + "p=0.10 # in w\n", + "t=24 # in h\n", + "Energy=p*t\n", + "print \"Energy = %0.2f kwh\"%Energy\n", + "cost=Energy*0.12\n", + "print \"Cost = %0.2f dollars\"%cost" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch18_1.ipynb b/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch18_1.ipynb new file mode 100644 index 00000000..941bf10e --- /dev/null +++ b/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch18_1.ipynb @@ -0,0 +1,284 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 18 : Direct current circuits" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 18.1 Page No: 597" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Solution a\n", + "Equivalent resistance = 18.00 ohm\n", + "Solution b\n", + "Current = 0.33 Amps\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "R1=2\n", + "R2=4\n", + "R3=5\n", + "R4=7\n", + "R_eq=R1+R2+R3+R4\n", + "v=6#in v\n", + "print \"Solution a\"\n", + "print \"Equivalent resistance = %0.2f ohm\"%R_eq\n", + "print \"Solution b\"\n", + "I=v/R_eq\n", + "print \"Current = %0.2f Amps\"%I" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example18.2 Page No: 599" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "solution a\n", + "Current = 6.00 amps\n", + "Current = 3.00 amps\n", + "Current = 2.00 amps\n", + "solution B\n", + "Power = 108.00 watt\n", + "Power = 54.00 watt\n", + "Power = 36.00 watt\n" + ] + } + ], + "source": [ + "delta_V=18#in volt\n", + "R1=3#in ohm\n", + "R2=6#in ohm\n", + "R3=9#in ohm\n", + "I1=delta_V/R1\n", + "I2=delta_V/R2\n", + "I3=delta_V/R3\n", + "print \"solution a\"\n", + "print \"Current = %0.2f amps\"%I1\n", + "print \"Current = %0.2f amps\"%I2\n", + "print \"Current = %0.2f amps\"%I3\n", + "P1=(I1**2)*R1\n", + "P2=(I2**2)*R2\n", + "P3=(I3**2)*R3\n", + "print \"solution B\"\n", + "print \"Power = %0.2f watt\"%P1\n", + "print \"Power = %0.2f watt\"%P2\n", + "print \"Power = %0.2f watt\"%P3" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 18.3 Page No: 602" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "solution b\n", + "Current = 3.00 amps\n" + ] + } + ], + "source": [ + "delta_Vac=42#in volt\n", + "R_eq=14#in ohm\n", + "I=delta_Vac/R_eq\n", + "print \"solution b\"\n", + "print \"Current = %0.2f amps\"%I" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 18.4 Page No: 605" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Current value I1 = -0.83, I2 = -0.53 & I3 = -0.30 amps\n" + ] + } + ], + "source": [ + "from numpy import mat\n", + "#formula used x=inv(a)*b\n", + "I=mat([[1 ,-1, -1],[-4, 0 ,-9],[0, -5, 9]])\n", + "V=mat([[0],[6],[0]])\n", + "X=(I**-1)\n", + "a=X*V\n", + "\n", + "print \"Current value I1 = %0.2f, I2 = %0.2f & I3 = %0.2f amps\"%(a[0],a[1],a[2])" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 18.5 Page No: 606" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Current value I1 = 2.00, I2 = -3.00 & I3 = -1.00 amps\n" + ] + } + ], + "source": [ + "from numpy import mat\n", + "#prob\n", + "#formula used x=inv(a)*b\n", + "I=mat([[8, 2, 0],[-3, 2, 0],[1, 1, -1]])\n", + "V=mat([[10],[-12],[0]])\n", + "X=I**-1\n", + "a=X*V\n", + "\n", + "print \"Current value I1 = %0.2f, I2 = %0.2f & I3 = %0.2f amps\"%(a[0],a[1],a[2])" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 18.6 Page No: 609" + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Constant of the circuit = 4.00 s\n", + "Charge = 6.00e-05 columb\n", + "Charge = 3.79e-05 columb when capacitance 63.2%\n" + ] + } + ], + "source": [ + "R=8*10**5#in ohms\n", + "C=5*10**-6#in Farad\n", + "t=R*C\n", + "print \"Constant of the circuit = %0.2f s\"%t\n", + "\n", + "Q=C*12\n", + "print \"Charge = %0.2e columb\"%Q\n", + "q=0.632*Q\n", + "print \"Charge = %0.2e columb when capacitance 63.2%%\"%q" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 18.7 Page No: 610" + ] + }, + { + "cell_type": "code", + "execution_count": 13, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "time = 1.39 s \n" + ] + } + ], + "source": [ + "from math import log\n", + "x=log(4)\n", + "print \"time = %0.2f s \"%x" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch19_1.ipynb b/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch19_1.ipynb new file mode 100644 index 00000000..3a586dc0 --- /dev/null +++ b/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch19_1.ipynb @@ -0,0 +1,333 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 19 : Magnetism" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 19.1 Page No: 631" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The force = 7.62e-19 Newton\n" + ] + } + ], + "source": [ + "q=1.6*10**-19#in columb\n", + "v=1*10**5#in m/s\n", + "B=55*10**-6#in T\n", + "F=q*v*B* 0.8660\n", + "print \"The force = %0.2e Newton\"%F" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 19.2 Page No: 632" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The force = 2.77e-12 Newton\n", + "Acceleration = 1.66e+15 m/s**2\n" + ] + } + ], + "source": [ + "q=1.6*10**-19#in columb\n", + "v=8*10**6#in m/s\n", + "B=2.5#in T\n", + "F=q*v*B* 0.8660\n", + "print \"The force = %0.2e Newton\"%F\n", + "m=1.67*10**-27\n", + "a=F/m\n", + "print \"Acceleration = %0.2e m/s**2\"%a" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 19.3 Page No: 635" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The maximaum force = 3.96e-02 Newton\n" + ] + } + ], + "source": [ + "l=36#in m\n", + "I=22#in A\n", + "B=0.50*10**-4#in T\n", + "F=B*I*l\n", + "print \"The maximaum force = %0.2e Newton\"%F" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 19.4 Page No: 637" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Torque = 0.39 N-m\n" + ] + } + ], + "source": [ + "from math import pi\n", + "A=pi*(0.5)*0.5#in m\n", + "I=2#in A\n", + "B=0.50#in T\n", + "T=B*I*A*0.5\n", + "print \"The Torque = %0.2f N-m\"%T" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 19.5 Page No: 640" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Velocity = 4.69e+06 m/s\n" + ] + } + ], + "source": [ + "q=1.6*10**-19\n", + "B=.35\n", + "r=14*10**-2#in m\n", + "m=1.67*10**-27#kg\n", + "v=(q*B*r)/m\n", + "print \"Velocity = %0.2e m/s\"%v" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 19.6 Page No: 641" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Radius of lighter istope = 0.10 m\n", + "Radius of heavier istope = 0.21 m\n", + "Distance of seperation = 0.21 m\n" + ] + } + ], + "source": [ + "q=1.6*10**-19\n", + "B=.10#in T\n", + "v=1*10**6#in m/s\n", + "r=14*10**-2#in m\n", + "m1=1.67*10**-27#in kg\n", + "m2=3.34*10**-27#in kg\n", + "r1=(m1*v)/(q*B)\n", + "r2=(m2*v)/(q*B)\n", + "x=(2*r2)-(2*r1)\n", + "print \"Radius of lighter istope = %0.2f m\"%r1\n", + "print \"Radius of heavier istope = %0.2f m\"%r2\n", + "print \"Distance of seperation = %0.2f m\"%x" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 19.7 Page No: 644" + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Magnetic field = 2.5e-04 T\n", + "Force = 6e-20 Newton\n" + ] + } + ], + "source": [ + "from math import pi\n", + "Uo=(4*pi*10**-7)\n", + "I=5#in A\n", + "r=4*10**-3\n", + "B=(Uo*I)/(2*pi*r)\n", + "print \"Magnetic field = %0.1e T\"%B\n", + "q=1.6*10**-19\n", + "v=1.5*10**3#in m/s\n", + "F=q*v*B\n", + "print \"Force = %0.e Newton\"%F" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 19.8 Page No: 646" + ] + }, + { + "cell_type": "code", + "execution_count": 14, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Current = 7.07 A\n" + ] + } + ], + "source": [ + "from math import pi, sqrt\n", + "mo=4*pi*10**-7#Tm/A\n", + "d=0.1#in m\n", + "x=1*10**-4#F/l\n", + "I=sqrt((x*2*pi*d)/mo)\n", + "print \"Current = %0.2f A\"%I" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 19.9 Page No: 649" + ] + }, + { + "cell_type": "code", + "execution_count": 15, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Magnetic field = 6.28e-04 T\n", + "Force = 1.88e-20 N\n" + ] + } + ], + "source": [ + "from math import pi\n", + "N=100#turns\n", + "l=.1#in m\n", + "n=N/l#in turns/m\n", + "mo=4*pi*10**-7#Tm/A\n", + "I=.5#in A\n", + "B=n*I*mo\n", + "q=1.6*10**-19#in c\n", + "v=375#in m/s\n", + "F=q*v*(B/2)\n", + "\n", + "print \"Magnetic field = %0.2e T\"%B\n", + "print \"Force = %0.2e N\"%F" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch20_1.ipynb b/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch20_1.ipynb new file mode 100644 index 00000000..475538f9 --- /dev/null +++ b/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch20_1.ipynb @@ -0,0 +1,316 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 20 : Induced voltages and inductance" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 20.1 Page No: 665" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Magnetic flux = 1.62e-04 T.m**2\n", + "Induced emf = 5.06e-03 volt\n" + ] + } + ], + "source": [ + "B=.5 # in T\n", + "A=3.24*10**-4 # in m**2\n", + "Flux=B*A\n", + "N=25\n", + "delta_t=.8\n", + "print \"Magnetic flux = %0.2e T.m**2\"%Flux\n", + "e=(N*Flux)/(delta_t)\n", + "print \"Induced emf = %0.2e volt\"%e" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 20.2 Page No: 667" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Induced emf = 0.45 volt\n" + ] + } + ], + "source": [ + "B=.6*10**-4 # in T\n", + "l=30\n", + "v=250 # in m/s\n", + "e=B*l*v\n", + "print \"Induced emf = %0.2f volt\"%e" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 20.3 Page No: 672" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Solution a\n", + "Induced emf = 0.25 volt\n", + "Solution b\n", + "Current = 0.50 A\n", + "Solution c\n", + "Power = 0.12 watt\n", + "Energy delivered = 0.12 J\n", + "Solution d\n", + "Force = 0.06 N\n" + ] + } + ], + "source": [ + "B=.25 # in T\n", + "l=.5\n", + "v=2 # in m/s\n", + "e=B*l*v\n", + "print \"Solution a\"\n", + "print \"Induced emf = %0.2f volt\"%e\n", + "R=.5 # in ohm\n", + "I=e/R\n", + "\n", + "print \"Solution b\"\n", + "print \"Current = %0.2f A\"%I\n", + "delta_v=.25\n", + "P=I*delta_v\n", + "print \"Solution c\"\n", + "print \"Power = %0.2f watt\"%P\n", + "t=1 # in s\n", + "w=P*t\n", + "print \"Energy delivered = %0.2f J\"%w\n", + " # Answer give for J in textbook is wrong\n", + "d=v*t\n", + "F=w/d\n", + "print \"Solution d\"\n", + "print \"Force = %0.2f N\"%F" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 20.5 Page No: 678" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Solution a\n", + "Induced emf = 135.72 volt\n", + "Solution b\n", + "Current = 11.31 A\n", + "Solution c\n", + "Emf in Volt 136*sinwt\n" + ] + } + ], + "source": [ + "from math import pi\n", + "f=60 # in Hz\n", + "w=2*pi*f\n", + "N=8\n", + "A=.09 # in m**2\n", + "B=.5 # in T\n", + "emf=N*A*B*w\n", + "print \"Solution a\"\n", + "print \"Induced emf = %0.2f volt\"%emf\n", + "R=12 # in ohm\n", + "I=emf/R\n", + "print \"Solution b\"\n", + "print \"Current = %0.2f A\"%I\n", + "\n", + "print \"Solution c\"\n", + "print \"Emf in Volt 136*sinwt\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 20.6 Page No: 680" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Solution a\n", + "Maximum Current = 12.00 A\n", + "Solution b\n", + "Current = 5.00 A\n" + ] + } + ], + "source": [ + "emf=120 # in Volt\n", + "R=10 # in Ohm\n", + "e_back=70\n", + "I=emf/R\n", + "print \"Solution a\"\n", + "print \"Maximum Current = %0.2f A\"%I\n", + "print \"Solution b\"\n", + "I=(emf-e_back)/R\n", + "print \"Current = %0.2f A\"%I" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 20.8 Page No: 684" + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Solution a\n", + "Inductance = 1.81e-04 H\n", + "Solution b\n", + "Emf = 9.05e-04 Volt\n" + ] + } + ], + "source": [ + "from math import pi\n", + "uo=4*pi*10**-7 # in m/A\n", + "N=300\n", + "A=4*10**-4 # in m**2\n", + "l=25*10**-2\n", + "L=(uo*N*N*A)/l\n", + "print \"Solution a\"\n", + "print \"Inductance = %0.2e H\"%L\n", + "delta_I=-5\n", + "delta_t=1\n", + "e=(-L*delta_I)/(delta_t)\n", + "print \"Solution b\"\n", + "print \"Emf = %0.2e Volt\"%e" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 20.9 Page No: 685" + ] + }, + { + "cell_type": "code", + "execution_count": 14, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Solution a\n", + "Time constant = 5.00e-03 s\n", + "Solution b\n", + "Current = 1.26 Amps\n" + ] + } + ], + "source": [ + "L=30*10**-3 # in Henry\n", + "R=6 # in Ohm\n", + "tou=L/R\n", + "print \"Solution a\"\n", + "print \"Time constant = %0.2e s\"%tou\n", + "\n", + "e=12\n", + "I=(0.632*e)/R\n", + "\n", + "\n", + "print \"Solution b\"\n", + "print \"Current = %0.2f Amps\"%I\n" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch21_1.ipynb b/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch21_1.ipynb new file mode 100644 index 00000000..6c83ccae --- /dev/null +++ b/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch21_1.ipynb @@ -0,0 +1,284 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 21 : Alternating current circuits and electromagnetic waves" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 21.1 Page No: 698" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Voltage = 141.42 V\n", + "Current = 1.41 Amps\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import sqrt\n", + "V_max=200#in V\n", + "V_rms=(V_max)/sqrt(2)\n", + "R=100#in ohm\n", + "I_rms=V_rms/R\n", + "print \"Voltage = %0.2f V\"%V_rms\n", + "print \"Current = %0.2f Amps\"%I_rms" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 21.2 Page No: 700" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Resistance = 331.56 ohm\n", + "Current = 0.45 Amps\n" + ] + } + ], + "source": [ + "C=8*10**-6\n", + "X_c=1/(377*C)\n", + "print \"Resistance = %0.2f ohm\"%X_c\n", + "I_rms=150/X_c\n", + "print \"Current = %0.2f Amps\"%I_rms" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 21.3 Page No: 702" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Resistance = 9.43 ohm\n", + "Current = 15.92 Amps\n" + ] + } + ], + "source": [ + "L=25*10**-3#In H\n", + "w=377\n", + "X_L=w*L#In ohm\n", + "print \"Resistance = %0.2f ohm\"%X_L\n", + "I_rms=150/X_L#In A\n", + "print \"Current = %0.2f Amps\"%I_rms" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 21.4 Page No: 706" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Impedence = 587.81 ohm\n", + "Current = 0.26 Amps\n", + "Angle = -64.83 degree\n", + "Voltage at Resistance = 63.80 Volt\n", + "Voltage at Inductance = 57.67 Volt\n", + "Voltage at Capacitance = 193.43 Volt\n" + ] + } + ], + "source": [ + "from math import atan, degrees, sqrt\n", + "R=250#in ohm\n", + "Xc=758#in ohm\n", + "Xl=226#in Ohm\n", + "X=Xl-Xc\n", + "V_max=150#in Volt\n", + "Z=sqrt(R**2+X**2)\n", + "I=V_max/Z\n", + "q=degrees(atan(X/R))\n", + "print \"Impedence = %0.2f ohm\"%Z\n", + "print \"Current = %0.2f Amps\"%I\n", + "print \"Angle = %0.2f degree\"%q\n", + "V_R=I*R\n", + "V_C=I*Xc\n", + "V_L=I*Xl\n", + "print \"Voltage at Resistance = %0.2f Volt\"%V_R\n", + "print \"Voltage at Inductance = %0.2f Volt\"%V_L\n", + "print \"Voltage at Capacitance = %0.2f Volt\"%V_C" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 21.5 Page No: 708" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Voltage = 106.07 V\n", + "Current = 0.18 Amps\n", + "Power = 8.15 watt\n" + ] + } + ], + "source": [ + "from math import sqrt,cos\n", + "V_max=150#in V\n", + "V_rms=(V_max)/sqrt(2)\n", + "I_max=.255#in ohm\n", + "I_rms=I_max/sqrt(2)\n", + "cos=.426\n", + "P=V_rms*I_rms*cos\n", + "print \"Voltage = %0.2f V\"%V_rms\n", + "print \"Current = %0.2f Amps\"%I_rms\n", + "print \"Power = %0.2f watt\"%P" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 21.6 Page No: 709" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Capacitance = 2e-06 Farad\n" + ] + } + ], + "source": [ + "L=20*10**-3#in H\n", + "C=1/(25*10**6*L)\n", + "print \"Capacitance = %0.e Farad\"%C" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 21.7 Page No: 711" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Solution a\n", + "Percentage of power lost = 0.02\n", + "Solution B\n", + "Percentage of power lost = 75.00\n" + ] + } + ], + "source": [ + "I1=100\n", + "v1=4*10**3\n", + "v2=2.40*10**5\n", + "I2=(I1*v1)/v2\n", + "R=30#in ohm\n", + "p_lost=I2*I2*R\n", + "P_output=I1*v1\n", + "p_per=(p_lost*100/P_output)\n", + "print \"Solution a\"\n", + "print \"Percentage of power lost = %0.2f\"%p_per\n", + "P_lost=I1*I1*R\n", + "per=(P_lost*100)/(4*10**5)\n", + "print \"Solution B\"\n", + "print \"Percentage of power lost = %0.2f\"%per" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch22_1.ipynb b/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch22_1.ipynb new file mode 100644 index 00000000..e4ba19da --- /dev/null +++ b/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch22_1.ipynb @@ -0,0 +1,169 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 22 : Reflection and refraction of light" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 22.2 Page No: 739" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Angle = 19.20 degree\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import sin, pi, degrees, asin\n", + "n1=1\n", + "n2=1.52\n", + "x=sin(pi/180*30)\n", + "theta_2=degrees(asin((n1*x)/n2))\n", + "print \"Angle = %0.2f degree\"%theta_2" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 22.3 Page No: 739" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Solution a\n", + "Velocity = 2.06e+08 m/s\n", + "Solution b\n", + "Wavelength in Fused quartz = 403.98 nm\n" + ] + } + ], + "source": [ + "print \"Solution a\"\n", + "c=3*10**8# Constant in m/s\n", + "n=1.458\n", + "v=c/n\n", + "print \"Velocity = %0.2e m/s\"%v\n", + "print \"Solution b\"\n", + "lambda_o=589#in nm\n", + "lambda_n=lambda_o/n\n", + "print \"Wavelength in Fused quartz = %0.2f nm\"%lambda_n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 22.5 Page No: 741" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Angle = 16.24 degree\n", + "Angle = 25.69 degree\n" + ] + } + ], + "source": [ + "from math import atan, degrees, asin\n", + "x=699#in micrometer(w-a)\n", + "t=1200 #in micrometer\n", + "b=x/2\n", + "theta_2=degrees(atan(b/t))\n", + "print \"Angle = %0.2f degree\"%theta_2\n", + "y=sin(pi/180*theta_2)\n", + "n1=1\n", + "n2=1.55\n", + "theta_1=degrees(asin((n2*y)/n1))\n", + "print \"Angle = %0.2f degree\"%theta_1" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 22.6 Page No: 744" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Angle(theta_c) = 48.75 degree\n" + ] + } + ], + "source": [ + "from math import asin, degrees\n", + "n1=1.33\n", + "n2=1\n", + "x=degrees(asin(n2/n1))\n", + "\n", + "print \"Angle(theta_c) = %0.2f degree\"%x" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch23_1.ipynb b/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch23_1.ipynb new file mode 100644 index 00000000..3f646e85 --- /dev/null +++ b/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch23_1.ipynb @@ -0,0 +1,367 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 23 : Mirrors and lenses" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 23.1 Page No: 760" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The hight = 0.90 m\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "AC= 1.8-.1#in m\n", + "AD=.5*AC\n", + "CF=.10#/in m\n", + "X=.5*CF#in m\n", + "FA=1.8#in m\n", + "d=FA-AD-X\n", + "print \"The hight = %0.2f m\"%d" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 23.2 Page No : 767" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "part a\n", + "The magnification when object is at 25cm : -0.67\n", + "part c\n", + "The magnification when object is at 5cm : 2.00\n" + ] + } + ], + "source": [ + "p=25#in cm\n", + "f=10#in cm\n", + "x=(1/f)-(1/p)\n", + "q=1/x\n", + "p=25\n", + "M=-(q/p)\n", + "print \"part a\"\n", + "print \"The magnification when object is at 25cm : %0.2f\"%M\n", + "p=5#in cm\n", + "f=10#in cm\n", + "x=(1/f)-(1/p)\n", + "q=1/x\n", + "p=5\n", + "M=-(q/p)\n", + "print \"part c\"\n", + "print \"The magnification when object is at 5cm : %0.2f\"%M" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 23.3 Page No: 768" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "part a\n", + "The position of final image = -5.71 cm\n", + "part b\n", + "The magnification : 0.23\n" + ] + } + ], + "source": [ + "p=20#in cm\n", + "f=-8#in cm\n", + "x=(1/f)-(1/p)\n", + "q=1/x\n", + "p=25\n", + "M=-(q/p)\n", + "print \"part a\"\n", + "print \"The position of final image = %0.2f cm\"%q\n", + "print \"part b\"\n", + "print \"The magnification : %0.2f\"%M" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 23.4 Page No: 769" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The focal length = 26.67 cm\n" + ] + } + ], + "source": [ + "p=40#in cm\n", + "q=-(2*p)\n", + "\n", + "x=(1/p)-(1/q)\n", + "f=1/x\n", + "print \"The focal length = %0.2f cm\"%f\n", + "#Answer given in book is wrong" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 23.5 Page No: 770" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The position of final image = -17.14 cm\n", + "The magnification when object = -1.29 cm\n", + "The Position of image = 2.57 cm\n" + ] + } + ], + "source": [ + "p=20#in cm\n", + "n1=1.5#in cm\n", + "n2=1#in cm\n", + "R=-30#in cm\n", + "x=(n2-n1)/R\n", + "y=n1/p\n", + "s=x-y\n", + "q=1/s\n", + "print \"The position of final image = %0.2f cm\"%q\n", + "M=(n1*q)/(n2*p)\n", + "print \"The magnification when object = %0.2f cm\"%M\n", + "h=2#in cm\n", + "h1=-M*h\n", + "print \"The Position of image = %0.2f cm\"%h1" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 23.7 Page No: 777" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "part a\n", + "The position of final image = 15.00 cm\n", + "The magnification : -0.50\n", + "part b\n", + "The position of final image = -10.00 cm\n", + "The magnification : 2.00\n" + ] + } + ], + "source": [ + "p=30#in cm\n", + "f=10#in cm\n", + "x=(1/f)-(1/p)\n", + "q=1/x\n", + "\n", + "M=-(q/p)\n", + "print \"part a\"\n", + "print \"The position of final image = %0.2f cm\"%q\n", + "print \"The magnification : %0.2f\"%M\n", + "p=5#in cm\n", + "f=10#in cm\n", + "x=(1/f)-(1/p)\n", + "q=1/x\n", + "M=-(q/p)\n", + "print \"part b\"\n", + "print \"The position of final image = %0.2f cm\"%q\n", + "print \"The magnification : %0.2f\"%M" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 23.8 Page No: 778" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "part a\n", + "The position of final image = -7.50 cm\n", + "The magnification : 0.25\n", + "part b\n", + "The position of final image = -5.00 cm\n", + "The magnification : 0.50\n", + "part c\n", + "The position of final image = -3.33 cm\n", + "The magnification : 0.67\n" + ] + } + ], + "source": [ + "p=30#in cm\n", + "f=-10#in cm\n", + "x=(1/f)-(1/p)\n", + "q=1/x\n", + "\n", + "M=-(q/p)\n", + "print \"part a\"\n", + "print \"The position of final image = %0.2f cm\"%q\n", + "print \"The magnification : %0.2f\"%M\n", + "p=10#in cm\n", + "f=-10#in cm\n", + "x=(1/f)-(1/p)\n", + "q=1/x\n", + "M=-(q/p)\n", + "print \"part b\"\n", + "print \"The position of final image = %0.2f cm\"%q\n", + "print \"The magnification : %0.2f\"%M\n", + "p=5#in cm\n", + "f=-10#in cm\n", + "x=(1/f)-(1/p)\n", + "q=1/x\n", + "M=-(q/p)\n", + "print \"part c\"\n", + "print \"The position of final image = %0.2f cm\"%q\n", + "print \"The magnification : %0.2f\"%M" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 23.9 Page No: 779" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The magnification : -0.67\n" + ] + } + ], + "source": [ + "p=30#in cm\n", + "f=10#in cm\n", + "x=(1/f)-(1/p)\n", + "q=1/x\n", + "\n", + "M1=-(q/p)\n", + "\n", + "p=5#in cm\n", + "f=20#in cm\n", + "x=(1/f)-(1/p)\n", + "q=1/x\n", + "\n", + "M2=-(q/p)\n", + "\n", + "\n", + "M=M1*M2\n", + "print \"The magnification : %0.2f\"%M" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch24_1.ipynb b/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch24_1.ipynb new file mode 100644 index 00000000..82f29c9f --- /dev/null +++ b/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch24_1.ipynb @@ -0,0 +1,233 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 24 : Wave optics" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1 Page No: 794" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(A) wavelength of light = 5.63e-07 meters\n", + "(B) Distance between adjacent fringes = 0.022 meters\n" + ] + } + ], + "source": [ + "L=1.2 # Seperation between screen and double-slit in meter\n", + "d=3*10**-5 #distance between the two slits\n", + "m=2 #second order bright fringe\n", + "Y=4.5*10**-2 #distance of second order bright fringe from centerline\n", + "#wavelength of light\n", + "lamda=(Y*d)/(m*L)\n", + "print \"(A) wavelength of light = %0.2e meters\"%lamda\n", + "#distance between adjacent bright fringes\n", + "#delta_Y=Y(m+1)-Ym\n", + "delta_Y=lamda*L/d\n", + "print \"(B) Distance between adjacent fringes = %0.3f meters\"%delta_Y" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2 Page No: 798" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Minimum thickness of soap bubble film = 113.16 nm is\n" + ] + } + ], + "source": [ + "n=1.33 #refractive index of soap bubble\n", + "lamda=602 #wavelength of light in nm\n", + "#for constructive interference we have 2nt=lamda/2\n", + "t=lamda/(4*n)\n", + "print \"Minimum thickness of soap bubble film = %0.2f nm is\"%t" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3 Page No: 799" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Minimum thickness of film = 95.17 nm is\n" + ] + } + ], + "source": [ + "n=1.45 #refractive index of silicon monoxide\n", + "lamda=552 #wavelength of light in nm\n", + "#for destructive interference we have condition for minimn thickness 2t=lamda/2n\n", + "t=lamda/(4*n)\n", + "print \"Minimum thickness of film = %0.2f nm is\"%t" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5 Page No: 801" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Pit depth in a CD = 121.88 nm\n" + ] + } + ], + "source": [ + "n=1.6 #refractive index of plastic transparent layer\n", + "lamda=780 #wavelength of laser light in nm\n", + "#for destructive interference we have condition for minimn thickness 2t=lamda/2n\n", + "t=lamda/(4*n)\n", + "print \"Pit depth in a CD = %0.2f nm\"%t" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6 Page No: 804" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " Position of first dark fringe = 3.87e-03 meters\n" + ] + } + ], + "source": [ + "lamda=580*10**-9 #wavelength of incident light in meter\n", + "a=0.30*10**-3 #slit width in meter\n", + "L=2 #distance of screen from slit in meters\n", + "#The first dark fringe corresponds to m=+1 or -1\n", + "m=1\n", + "sin_theta=m*lamda/a\n", + "#From fig 24.16 tan_theta=y/L and since theta is very small we have sin_theta=tan_theta hence sin_theta=y/L\n", + "y=L*sin_theta \n", + "print \" Position of first dark fringe = %0.2e meters\"%y" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7 Page No: 808" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Angle in degrees at which first order maxima is observed : 22.32\n", + "Angle in degrees at which second order maxima is observed : 49.43\n", + "for higher order number of diffraction the the solutions are non realistic\n" + ] + } + ], + "source": [ + "from math import asin, degrees\n", + "lamda=632.8 #wavelength of monochromatic light from helium-neon laser in meter\n", + "a=6000 #lines in diffraction grating per cm\n", + "d=10**7/a#slit seperation in nm\n", + "#for the first order maximum we have m=1\n", + "sin_theta1=lamda/d\n", + "theta1=degrees(asin(sin_theta1))\n", + "print \"Angle in degrees at which first order maxima is observed : %0.2f\"%theta1\n", + "#for the second order maximum we have m=2\n", + "sin_theta2=2*lamda/d\n", + "theta2=degrees(asin(sin_theta2))\n", + "print \"Angle in degrees at which second order maxima is observed : %0.2f\"%theta2\n", + "print \"for higher order number of diffraction the the solutions are non realistic\"" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch25_1.ipynb b/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch25_1.ipynb new file mode 100644 index 00000000..918e8b63 --- /dev/null +++ b/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch25_1.ipynb @@ -0,0 +1,291 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 25 : Optical Instruments" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1 Page No: 827" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "a) focal length f = 50.00 cm\n", + "b) Power of the lens = 2.00 diopters\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "q=-50 # Near point of an eye in cm\n", + "p=25 #object location in cm\n", + "#a) focal length calculation\n", + "#Using Thin Lens equation 1/f=((1/p)+(1/q))\n", + "f=p*q/(p+q)\n", + "print \"a) focal length f = %0.2f cm\"%f\n", + "#b) power of the lens\n", + "f1=50*10**-2# focal length in meters\n", + "P=1/f1\n", + "print \"b) Power of the lens = %0.2f diopters\"%P" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3 Page No: 830" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "a) Maximum angular magnification of the lens : 3.50\n", + "Angular Magnification of lens when eye is relaxed : 2.50\n" + ] + } + ], + "source": [ + "f=10 # focal length in cm\n", + "#a)Maximum angular magnification\n", + "M_max=1+(25/f)\n", + "print \"a) Maximum angular magnification of the lens : %0.2f\"%M_max\n", + "m=25/f\n", + "print \"Angular Magnification of lens when eye is relaxed : %0.2f\"%m" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4 Page No: 832" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Magnification of microscope with two long focal lengths : -45.00\n", + "Magnification of microscope with a combination of 20 mm objective and 2.5 cm eyepiece : -90.00\n", + "Magnification of microscope with a combination of 20 mm objective and 2.5 cm eyepiece : -450.00 \n", + "Possible magnification of microscope with two short focal lengths : -900.00\n" + ] + } + ], + "source": [ + "#interchangeable objectives\n", + "f1=2 # focal length in cm\n", + "f2=0.2 #focal length in cm\n", + "#data of two eye pieces\n", + "f3=5 #focal length in cm\n", + "f4=2.5 #focal length in cm\n", + "L=18 # length of microscope\n", + "#Calculation of magnification for four combinations of lens\n", + "#magnification of compound microscope m =-(L/fo)*(25cm/fe) where fo is shortest focal length compared to fe\n", + "#combination of two long focal lengths\n", + "m1=-(L/f1)*(25/f3)\n", + "print \"Magnification of microscope with two long focal lengths : %0.2f\"%m1\n", + "#combination of 20 mm objective and 2.5 cm eyepiece\n", + "m2=-(L/f1)*(25/f4)\n", + "print \"Magnification of microscope with a combination of 20 mm objective and 2.5 cm eyepiece : %0.2f\"%m2\n", + "#combination of 2 mm objective and 5 cm eyepiece\n", + "m3=-(L/f2)*(25/f3)\n", + "print \"Magnification of microscope with a combination of 20 mm objective and 2.5 cm eyepiece : %0.2f \"%m3\n", + "#combination of two short focal lengths\n", + "m4=-(L/f2)*(25/f4)\n", + "print \"Possible magnification of microscope with two short focal lengths : %0.2f\"%m4" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5 Page No: 834" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Angular magnification of the telescope : 83.33\n" + ] + } + ], + "source": [ + "d=8 #diameter of objective mirror of reflecting telescope in inches\n", + "fo=1500 #focal length of objective mirror of reflecting telescope in mm\n", + "fe=18 #focal length of eyepiece\n", + "m=fo/fe\n", + "print \"Angular magnification of the telescope : %0.2f\"%m" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6 Page No: 837" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "a) Limiting angle of resolution in radians : 7.98e-07\n", + "b) Maximum limit of resolution for the microscope in radians : 5.42e-07\n", + "c) Limiting angle of resolution for the microscope when water filled the space b/w the object and objective in radians : 6.00e-07\n" + ] + } + ], + "source": [ + "l=589*10**-9 #Wavelength of sodium light m\n", + "d=90*10**-2 #diameter of the aperture in m \n", + "L=400*10**-9 #Wavelength of desirable Visble light\n", + "n=1.33 #refractive index of water\n", + "#a) Calculation of limiting angle of resolution\n", + "#Limiting angle of resolution of the circular aperture is Theta_min=1.22*(l/d)\n", + "Theta_min1=1.22*(l/d)\n", + "print \"a) Limiting angle of resolution in radians : %0.2e\"%Theta_min1\n", + "#b) Calculation of maximum limit of resolution for the microscope\n", + "Theta_min2=1.22*(L/d)\n", + "print \"b) Maximum limit of resolution for the microscope in radians : %0.2e\"%Theta_min2\n", + "#c)Effect of water b/w the object and objective on resolving power of microscope\n", + "lw=l/n\n", + "Theta_min3=1.22*(lw/d)\n", + "print \"c) Limiting angle of resolution for the microscope when water filled the space b/w the object and objective in radians : %0.2e\"%Theta_min3" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7 Page No: 838" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Magnification of telescope A is : 166.67\n", + "Magnification of telescope B is : 50.00\n" + ] + } + ], + "source": [ + "f1=1000# focal length of objective of telescope A in mm\n", + "f2=1250# focal length of objective of telescope B in mm\n", + "f3=6# focal length of eyepiece of telescope A in mm\n", + "f4=25# focal length of eyepiece of telescope Bin mm\n", + "#C) Calculation of magnification of the telescope\n", + "m_A=f1/f3\n", + "m_B=f2/f4\n", + "print \"Magnification of telescope A is : %0.2f\"%m_A\n", + "print \"Magnification of telescope B is : %0.2f\"%m_B" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8 Page No: 839" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "a) Resolving poer of grating inorder to distinguish the wavelengths = 998.31\n", + "b) No.of lines of the grating illuminated to resolve the lines in the second order spectrum are 499\n" + ] + } + ], + "source": [ + "L1=589 # wavelength of first bright line in sodium spectrum in nm\n", + "L2=589.59 # wavelength of second bright line in sodium spectrum in nm\n", + "m=2 # order of the spectrum\n", + "delta_L=L2-L1\n", + "R=L1/delta_L\n", + "print \"a) Resolving poer of grating inorder to distinguish the wavelengths = %0.2f\"% R\n", + "N=R/m\n", + "print \"b) No.of lines of the grating illuminated to resolve the lines in the second order spectrum are %d\"%N" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch26_1.ipynb b/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch26_1.ipynb new file mode 100644 index 00000000..461f6400 --- /dev/null +++ b/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch26_1.ipynb @@ -0,0 +1,350 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 26 : Relativity" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example1 Page No: 855" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Period of the pendulum w.r.t to observer = 9.61 \n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import sqrt\n", + "Tp=3 #proper time in sec\n", + "c=3*10**8 #velocity of light in m/sec\n", + "v=0.95*c\n", + "gamma=1/sqrt(1-(v**2/c**2))\n", + "T=gamma*Tp\n", + "print \"Period of the pendulum w.r.t to observer = %.2f \"%T" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example2 Page No: 857" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Length of spaceship measured by moving observer = 16.93 meters\n" + ] + } + ], + "source": [ + "from math import sqrt\n", + "Lp=120 # length of space ship in meters\n", + "c=3*10**8 #velocity of light in m/sec\n", + "v=0.99*c\n", + "gamma=1/sqrt(1-(v**2/c**2))\n", + "L=Lp/gamma\n", + "print \"Length of spaceship measured by moving observer = %0.2f meters\"%L" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example3 Page No: 859" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Distance from spaceship to the groung measured by an observer in spaceship = 105.75 meters\n" + ] + } + ], + "source": [ + "from math import sqrt\n", + "Lp=435 # length of space ship in meters\n", + "c=3*10**8 #velocity of light in m/sec\n", + "v=0.970*c\n", + "gamma=1/sqrt(1-(v**2/c**2))\n", + "L=Lp/gamma\n", + "print \"Distance from spaceship to the groung measured by an observer in spaceship = %0.2f meters\"%L" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example4 Page No: 861" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The observer sees the horizontal dimension of the spaceship gets contracted to a length of 16.24 meters\n" + ] + } + ], + "source": [ + "from math import sqrt\n", + "c=3*10**8 #velocity of light in m/sec\n", + "#when the spaceship is at rest\n", + "x=52 # diatance in x direction in meters\n", + "y=25 #measurement in y direction\n", + "v=0.95*c\n", + "#when the spaceship moves to an observer at rest only x dimension looks contracted\n", + "gamma=1/sqrt(1-(v**2/c**2))\n", + "L=x/gamma\n", + "print \"The observer sees the horizontal dimension of the spaceship gets contracted to a length of %0.2f meters\"%L" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example5 Page No: 862" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "relativistic momentum = 3.10e-22 kg.m/s\n", + "classical momentum = 2.05e-22 kg.m/s\n", + "the relativistic result is 51 percent greater than classical result\n" + ] + } + ], + "source": [ + "from math import sqrt\n", + "c=3*10**8 #velocity of light in m/sec\n", + "m=9.11*10**-31 #mass of electron in kg\n", + "v=0.75*c\n", + "gamma=1/sqrt(1-(v**2/c**2))\n", + "#relativistic momentum\n", + "p=m*v*gamma\n", + "print \"relativistic momentum = %0.2e kg.m/s\"%p\n", + "#classical approach\n", + "P=m*v\n", + "print \"classical momentum = %0.2e kg.m/s\"%P\n", + "Z=(p-P)*100/P\n", + "print \"the relativistic result is %d percent greater than classical result\"%Z" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example6 Page No: 864" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "velocity of light w.r.t stationary observer = 3.00e+08 m/sec\n" + ] + } + ], + "source": [ + "c=3*10**8 #velocity of light in m/sec\n", + "Vmo=0.80*c # velocity of motocycle w.r.t stationary observer \n", + "Vlm=c # velocity of motocycle w.r.t motorcycle\n", + "#velocity of light w.r.t stationary observer \n", + "Vlo=(Vlm+Vmo)/(1+(Vlm*Vmo)/c**2)\n", + "print \"velocity of light w.r.t stationary observer = %0.2e m/sec\"%Vlo" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example7 Page No: 865" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The energy equivalent of baseball = 4.50e+16 joules\n" + ] + } + ], + "source": [ + "c=3*10**8 #velocity of light in m/sec\n", + "m=0.50 #mass of baseball in kg\n", + "E=m*c**2\n", + "print \"The energy equivalent of baseball = %0.2e joules\"%E" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8 Page No: 866" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "total energy of an electron = 0.97 Mev\n", + "Kinetic energy of electron = 0.46 Mev\n" + ] + } + ], + "source": [ + "from math import sqrt\n", + "c=3*10**8 #velocity of light in m/sec\n", + "m=0.511 #rest energy of electron in Mev\n", + "v=0.85*c\n", + "gamma=1/sqrt(1-(v**2/c**2))\n", + "E=(m)*gamma\n", + "print \"total energy of an electron = %0.2f Mev\"%E\n", + "K=E-m\n", + "print \"Kinetic energy of electron = %0.2f Mev\"%K" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9 Page No: 867" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "a) Kinetic energy released in fission = 200.62 Mev\n", + " Speed of Barium fragment = 1.66e+07 Mev\n", + " Speed of krypton fragment = 2.05e+07 Mev\n" + ] + } + ], + "source": [ + "from math import sqrt\n", + "m_n=1.008665 #mass of neutron in amu\n", + "m_U=235.043924 #atomic mass of uranium in amu\n", + "m_Ba=140.903496 #atomic mass of barium in amu\n", + "m_Kr=91.907720 #atomic mass of krypton in amu\n", + "c=3*10**8 # velocity of light in m/s\n", + "#a) Kinetic energy released in fission of uranium\n", + "KE_final_=((m_n+m_U)-(m_Ba+m_Kr+(3*m_n)))*c**2\n", + "#1 amu = 931.494 Mev/c**2\n", + "KE_final=KE_final_*931.494/c**2\n", + "print \"a) Kinetic energy released in fission = %0.2f Mev\"%KE_final\n", + "#b) velocities of barium and krypton\n", + "#E=mc2/sqrt(1-v2/c2)\n", + "KE_Ba=KE_final\n", + "m_Ba_=m_Ba*931.494/c**2 # mass of barium in Mev\n", + "E_Ba=KE_Ba+m_Ba_*c**2\n", + "V_Ba=(sqrt(1-(((m_Ba_*c**2)**2)/E_Ba**2)))*c\n", + "print \" Speed of Barium fragment = %0.2e Mev\"%V_Ba\n", + "KE_Kr=KE_final\n", + "m_Kr_=m_Kr*931.494/c**2 # mass of krypton in Mev\n", + "E_Kr=KE_Kr+m_Kr_*c**2\n", + "V_Kr=(sqrt(1-((m_Kr_*c**2)**2)/E_Kr**2))*c\n", + "print \" Speed of krypton fragment = %0.2e Mev\"%V_Kr\n", + "#The difference in answer is because of round off" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch27_1.ipynb b/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch27_1.ipynb new file mode 100644 index 00000000..cc3bdd59 --- /dev/null +++ b/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch27_1.ipynb @@ -0,0 +1,401 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 27 : Quantum physics" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1 Page No: 874" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Wavelength at which radiation emitted from the skin reaches its peak = 9.41e-06 meters\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "T=35 #Temperature of the skin in celsius\n", + "T1=T+273 #Temperature in kelvin\n", + "#From Wien's displacement law \n", + "Lambda_max=(0.2898*10**-2)/T1\n", + "print \"Wavelength at which radiation emitted from the skin reaches its peak = %0.2e meters\"%Lambda_max" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2 Page No: 878" + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "a) Total energy of Simple harmonic oscillator with given amplitude = 2.00 Joules\n", + " Frequency of oscillation = 0.56 Hertz\n", + "b) Quantum number for the given macroscopic system : 5.36e+33\n", + "c) Energy carried away by a one-quantum charge = 3.73e-34 joules\n" + ] + } + ], + "source": [ + "from math import sqrt, pi\n", + "m=2 # mass of the object in Kg\n", + "k=25 #force constant of spring in N/m\n", + "A=0.4 #Amplitude of Simple harmonic oscillation by spring in meters\n", + "h=6.63*10**-34#js\n", + "#a) Total energy and frequency of SHO calculation\n", + "E=(1/2)*k*A**2\n", + "f=(1/(2*pi))*sqrt(k/m)\n", + "print \"a) Total energy of Simple harmonic oscillator with given amplitude = %0.2f Joules\"%E\n", + "print \" Frequency of oscillation = %0.2f Hertz\"%f\n", + "#b) Calculation of quantum number for the system\n", + "n=E/(h*f)\n", + "print \"b) Quantum number for the given macroscopic system : %0.2e\"%n\n", + "#c) Calculation of energy carried away in a quantum charge\n", + "delta_E=h*f\n", + "print \"c) Energy carried away by a one-quantum charge = %0.2e joules\"%delta_E" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3 Page No: 879" + ] + }, + { + "cell_type": "code", + "execution_count": 14, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Energy carried by a photon with the given frequency = 3.98e-19 Joules\n" + ] + } + ], + "source": [ + "f=6*10**14 #frequency of yellow light in hertz\n", + "h=6.63*10**-34 #plancks constant J.s\n", + "E=h*f\n", + "print \"Energy carried by a photon with the given frequency = %0.2e Joules\"%E" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4 Page No: 882" + ] + }, + { + "cell_type": "code", + "execution_count": 16, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "a) Maximum Kinetic energy of th eejected photoelectrons = 1.68 ev is\n", + "b) Cut off wavelength for sodium = 5.05e-07 meters\n" + ] + } + ], + "source": [ + "l=0.3*10**-6 #wavelength of light in meters\n", + "W=2.46 #work function for sodium in ev\n", + "c=3*10**8 #velocity of light in m/s\n", + "h=6.63*10**-34#js\n", + "#a) Maximum KE of the ejected photoelectrons\n", + "E=(h*c/l)/(1.6*10**-19) #energy of each photon of th eilluminating light beam in ev\n", + "KE_max=E-W\n", + "print \"a) Maximum Kinetic energy of th eejected photoelectrons = %0.2f ev is\"%KE_max\n", + "#b) Cut off wavelength for sodium \n", + "W1=W*1.6*10**-19\n", + "lc=h*c/W1\n", + "print \"b) Cut off wavelength for sodium = %0.2e meters\"%lc" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5 Page No: 885" + ] + }, + { + "cell_type": "code", + "execution_count": 18, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Minimum wavelength produced = 1.24e-11 meters\n" + ] + } + ], + "source": [ + "V=10**5 #potential difference in Volts\n", + "h=6.63*10**-34 # plancks constant in J.s\n", + "c=3*10**8# velocity of light in m/s\n", + "e=1.6*10**-19# elelctronic charge in coulombs\n", + "L_min=(h*c)/(e*V)\n", + "print \"Minimum wavelength produced = %0.2e meters\"%L_min" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6 Page No: 886" + ] + }, + { + "cell_type": "code", + "execution_count": 20, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Grazing angle at first order of interference = 6.40 degree\n", + "Grazing angle at third order of interference = 19.54 degree\n" + ] + } + ], + "source": [ + "from math import asin, degrees\n", + "d=0.314 #spacing between certain planes in a crystal of calcite in nm\n", + "l=0.070 #wavelength of X-rays in nm\n", + "m=1# first order of interference\n", + "theta1=degrees(asin((m*l)/(2*d)))\n", + "print \"Grazing angle at first order of interference = %0.2f degree\"%theta1\n", + "m=3 #third order of interference\n", + "theta2=degrees(asin((m*l)/(2*d)))\n", + "print \"Grazing angle at third order of interference = %0.2f degree\"%theta2" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7 Page No: 887" + ] + }, + { + "cell_type": "code", + "execution_count": 23, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Wavelength of the scattered X-rays at the given angle in 0.20 nm\n" + ] + } + ], + "source": [ + "from math import pi, cos\n", + "Lo=0.200000 #wavelength of X-rays in nm\n", + "h=6.63*10**-34 #in J.s\n", + "m_e=9.11*10**-31 # in Kg\n", + "c=3*10**8 #in m/s\n", + "theta=45 #in degrees\n", + "#wavelength is represented by d\n", + "delta_L=(h/(m_e*c))*(1-cos(pi/180*theta))\n", + "L=delta_L+Lo\n", + "print \"Wavelength of the scattered X-rays at the given angle in %.2f nm\"%L\n", + "\n", + "#Answer given in textbook is wrong" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8 Page No: 887" + ] + }, + { + "cell_type": "code", + "execution_count": 25, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "de Broglie wavelength for an electron = 7.28e-11 meters\n" + ] + } + ], + "source": [ + "h=6.63*10**-34 #in J.s\n", + "m_e=9.11*10**-31 # in Kg\n", + "v=1*10**7 #in m/s\n", + "lamda=h/(m_e*v)\n", + "print \"de Broglie wavelength for an electron = %0.2e meters\"%lamda" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9 Page No: 888" + ] + }, + { + "cell_type": "code", + "execution_count": 27, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "de Broglie wavelength of the ball = 1.14e-34 meters\n" + ] + } + ], + "source": [ + "h=6.63*10**-34 #in J.s\n", + "m=0.145 # in Kg\n", + "v=40 #in m/s\n", + "lamda=h/(m*v)\n", + "print \"de Broglie wavelength of the ball = %0.2e meters\"%lamda" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 10 Page No: 889" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Uncertainity in position of electron = 3.86e-06 Meters\n" + ] + } + ], + "source": [ + "from math import pi\n", + "h=6.63*10**-34#js\n", + "v=5*10**3 #speed of the electron in m/s\n", + "m_e=9.11*10**-31 # mass of electron in Kg\n", + "p=m_e*v\n", + "delta_p=0.00300*p\n", + "#Uncertainity principle states delta_x*delta_p >=h/(4*pi)\n", + "delta_x=h/(4*pi*delta_p)\n", + "print \"Uncertainity in position of electron = %0.2e Meters\"%delta_x" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 11 Page No: 889" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " Minimum uncertainity in energy of the excited states = 5.28e-27 Joules\n" + ] + } + ], + "source": [ + "from math import pi\n", + "h=6.63*10**-34 # plancks constant in J.s\n", + "delta_t=1.00*10**-8 # Average time that an ellectron exists in the excited states in sec\n", + "delta_E=h/(4*pi*delta_t)\n", + "print \" Minimum uncertainity in energy of the excited states = %0.2e Joules\"%delta_E" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch28_1.ipynb b/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch28_1.ipynb new file mode 100644 index 00000000..941b57e6 --- /dev/null +++ b/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch28_1.ipynb @@ -0,0 +1,197 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 28 : Atomic Physics" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1 Page No: 897" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Wavelength of the emitted photon = 1.22e-07 meters\n", + "frequency of the emitted photon = 2.47e+15 meters\n" + ] + } + ], + "source": [ + "RH=1.097*10**7 #Rydberg constant in per meter\n", + "lamda=4/(3*RH)\n", + "c=3*10**8 # m/sec\n", + "f=c/lamda\n", + "print \"Wavelength of the emitted photon = %0.2e meters\"%lamda\n", + "print \"frequency of the emitted photon = %0.2e meters\"%f" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2 Page No: 898" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "longest wavelength that photon emmited = 6.56e-07 meters\n", + "Energy emmited by the photon = 3.03e-19 Joules\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "RH=1.097*10**7 #Rydberg constant in per meter\n", + "h=6.626*10**-34 #plancks constant in j.s\n", + "c=3*10**8 # velocity of light in m/s\n", + "nf=2 #quantum number\n", + "ni=3# quantum number\n", + "#assuming k=1/lamda\n", + "k=RH*((1/nf**2-1/ni**2))\n", + "lamda=1/k\n", + "print \"longest wavelength that photon emmited = %0.2e meters\"%lamda\n", + "E_photon=h*c/lamda\n", + "print \"Energy emmited by the photon = %0.2e Joules\"%E_photon" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3 Page No: 901" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "a) Energy of the atom in ground state = -54.40 eV\n", + "b) Radius of the ground state orbit = 0.000 nm\n" + ] + } + ], + "source": [ + "Z=2 #atomic number of helium\n", + "n=1 #principal quantum number\n", + "E=-Z**2*13.6/n**2\n", + "print \"a) Energy of the atom in ground state = %0.2f eV\"%E\n", + "r=(n**2/Z)*0.0529#in nm\n", + "print \"b) Radius of the ground state orbit = %0.3f nm\"%r" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4 Page No: 906" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Energy of the states with quantum number 2 = -3.40 ev\n" + ] + } + ], + "source": [ + "n=2# principal quantum number \n", + "E=-13.6/n**2\n", + "print \"Energy of the states with quantum number 2 = %0.2f ev\"%E" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6 Page No: 906" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Energy of the characteristic emiited from tungsten target when electron drops from M shell to K shell = 6.61e+04 ev\n" + ] + } + ], + "source": [ + "Z=74 #atomic number of tungsten\n", + "Eo=13.6 #ground state enenrgy in ev\n", + "E_K=-(Z-1)**2*(13.6) #Energy of the electron in K shell\n", + "n=3\n", + "Z_eff=Z-n**2\n", + "E3=Eo/n**2\n", + "E_M=-Z_eff**2*E3\n", + "E=E_M-E_K\n", + "print \"Energy of the characteristic emiited from tungsten target when electron drops from M shell to K shell = %0.2e ev\"%E\n", + "#Difference in answer is because of roundoff" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch29_1.ipynb b/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch29_1.ipynb new file mode 100644 index 00000000..e3458b6c --- /dev/null +++ b/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch29_1.ipynb @@ -0,0 +1,263 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 29 : Nuclear Physics" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1 Page No: 916" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Nuclear density = 2.31e+17 kg/m3\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import pi\n", + "m=1.67*10**-27 #mass of nucleus in kg\n", + "ro=1.2*10**-15 #in meter\n", + "p=(3*m)/(4*pi*(ro)**3)\n", + "print \"Nuclear density = %0.2e kg/m3\"%p" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2 Page No: 920" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Binding energy of Deuteron = 2.22 Mev\n" + ] + } + ], + "source": [ + "mp=1.007825 #in u\n", + "mn=1.008665 #in u\n", + "md=2.014102 #in u\n", + "u=931.494 #Mev\n", + "M=mp+mn\n", + "delta_m=(M-md) #in u\n", + "E=delta_m*u\n", + "print \"Binding energy of Deuteron = %0.2f Mev\"%E" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3 Page No: 922" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Activity or decay rate at t=0 = 1.11e-05 Ci\n" + ] + } + ], + "source": [ + "No=3*10**16 #no.of radioactive nuclei present at t=0\n", + "t_half=1.6*10**3 #years\n", + "T_half=t_half*3.16*10**7 #in sec\n", + "d=0.693/T_half\n", + "R_o=d*No # decays/s\n", + "Ci=3.7*10**10\n", + "Ro=R_o/Ci\n", + "print \"Activity or decay rate at t=0 = %0.2e Ci\"%Ro" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4 Page No: 923" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "a) No.of atoms remaining after 12 days : 45612654\n", + "Initial activity of the radon sample = 837.68 decay/sec\n" + ] + } + ], + "source": [ + "from math import exp\n", + "T_half=3.83 #half life time of Radon in days\n", + "No=4*10**8 #Initial No .of Radon atoms \n", + "lamda=0.693/T_half # in days\n", + "t=12 \n", + "N=No*exp(-(lamda*t))\n", + "print \"a) No.of atoms remaining after 12 days : %0.f\"%N\n", + "lamda_=lamda/(8.64*10**4)\n", + "R=lamda_*No\n", + "print \"Initial activity of the radon sample = %0.2f decay/sec\"%R" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5 Page No: 925" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Energy liberated = 4.87 Mev\n" + ] + } + ], + "source": [ + "m_d=222.017571 #mass of daughter nuclei in atomic units\n", + "m_alpha=4.002602 #mass of alpha particle in atomic units\n", + "M_p=226.025402 #mass of parent nuclei in atomic units\n", + "m=m_d+m_alpha\n", + "delta_m=(M_p-m)\n", + "E=delta_m*931.494\n", + "print \"Energy liberated = %0.2f Mev\"%E" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6 Page No: 927" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Energy released in beta decay = 0.156 Mev\n" + ] + } + ], + "source": [ + "M_C=14.003242 #mass of carbon in atomic mass units\n", + "M_N=14.003074 #mass of nitogen in atomic mass units\n", + "delta_M=M_C-M_N\n", + "E=delta_M*(931.494)\n", + "print \"Energy released in beta decay = %0.3f Mev\"%E" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7 Page No: 928" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Age of the skeleton = 10915.43 years\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import log\n", + "T_half=3.01*10**9 #half life time in min\n", + "lamda=0.693/T_half\n", + "R=200 # in decay/min\n", + "R0_=15 #decay rate in decay/min.g\n", + "m=50 #weight of carbon\n", + "R0=R0_*m #in decay/min\n", + "t1=-(log(R/R0)/lamda) #im min\n", + "t=t1/525949\n", + "print \"Age of the skeleton = %0.2f years\"%t" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch30_1.ipynb b/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch30_1.ipynb new file mode 100644 index 00000000..3ab7db90 --- /dev/null +++ b/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch30_1.ipynb @@ -0,0 +1,107 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 30 : Nuclear energy and elementary particles" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2 Page No: 943" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Disintegration energy = 5.33e+26 Mev is\n", + "or = 2.37e+07 KWh\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "Q=208 #disintegration energy per event in Mev\n", + "m=1*10**3 #mass of uranium\n", + "A=235 #mass number or uranium in g/mol\n", + "a=6.02*10**23 #avagadro number nuclei/mol\n", + "N=(a/A)*m #nuclei\n", + "E=N*Q\n", + "P=E*4.45*10**-20\n", + "print \"Disintegration energy = %0.2e Mev is\"%E\n", + "print \"or = %0.2e KWh\"%P" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3 Page No: 947" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " Energy release in deuterium-deuterium reaction = 4.03 Mev\n" + ] + } + ], + "source": [ + "m1=2.014102 # mass of deuterium in atomic mass unit\n", + "m2=3.016049 #mass of tritium in atomic mass unit\n", + "m3=1.007825 # mass of hydrogen in atomic mass unit\n", + "#referring to the deuterium-deuterium reaction\n", + "#mass before reaction\n", + "M1=2*m1\n", + "#mass after reaction\n", + "M2=m2+m3\n", + "#excessive mass\n", + "m=M1-M2\n", + "#converting mass into energy\n", + "#1 u = 931.494 Mev\n", + "E=m*931.494\n", + "print \" Energy release in deuterium-deuterium reaction = %0.2f Mev\"%E" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/screenshots/1RefractionOfLaserLight_1.png b/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/screenshots/1RefractionOfLaserLight_1.png Binary files differnew file mode 100644 index 00000000..a2cdf1a4 --- /dev/null +++ b/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/screenshots/1RefractionOfLaserLight_1.png diff --git a/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/screenshots/2PropertiesOfImage_1.png b/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/screenshots/2PropertiesOfImage_1.png Binary files differnew file mode 100644 index 00000000..499e2c33 --- /dev/null +++ b/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/screenshots/2PropertiesOfImage_1.png diff --git a/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/screenshots/3PositionOf1DarkFringe_1.png b/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/screenshots/3PositionOf1DarkFringe_1.png Binary files differnew file mode 100644 index 00000000..b0d4c88b --- /dev/null +++ b/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/screenshots/3PositionOf1DarkFringe_1.png diff --git a/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/Chapter10_1.ipynb b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/Chapter10_1.ipynb new file mode 100644 index 00000000..c3f165d1 --- /dev/null +++ b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/Chapter10_1.ipynb @@ -0,0 +1,938 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:08446d35c254f7719d0bb2f900fc25303d15f487208d5c6fccc1dc48e1acf8aa"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 10 Magnets and Earth's Magnetism"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 10.1 Page no 558"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "F=0.8*10**-3*9.8 #N\n",
+ "d=0.1 #m\n",
+ "u=10**-7\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "m=math.sqrt(F*d**2/(u*5))\n",
+ "m1=5*m\n",
+ "\n",
+ "#Result\n",
+ "print\"Strength of pole M1 is\", round(m,2),\"Am\"\n",
+ "print\"Strength of pole M2 is\",round(m1,1),\"Am\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Strength of pole M1 is 12.52 Am\n",
+ "Strength of pole M2 is 62.6 Am\n"
+ ]
+ }
+ ],
+ "prompt_number": 11
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 10.2 Page no 559"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "F=14.4*10**-4 #N\n",
+ "d=0.05 #m\n",
+ "F1=1.6*10**-4\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "u=4*math.pi*10**-7\n",
+ "m=math.sqrt((F*4*math.pi*d**2)/u)\n",
+ "d1=1/(math.sqrt((F1*4*math.pi)/(u*m**2)))\n",
+ "\n",
+ "#Result\n",
+ "print \"Distance is\",d1,\"m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Distance is 0.15 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 19
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 10.5 Page no 560"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#given\n",
+ "M=8\n",
+ "d=0.2\n",
+ "\n",
+ "#Calculation\n",
+ "B=u*2*M/(4*math.pi*d**3)\n",
+ "Beqa=B/2.0\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Magnetic induction at axial point\", B*10**4,\"*10**-4 T\"\n",
+ "print\"(ii) Magnetic induction at equatorial point is\",Beqa*10**4,\"*10**-4 T\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Magnetic induction at axial point 2.0 *10**-4 T\n",
+ "(ii) Magnetic induction at equatorial point is 1.0 *10**-4 T\n"
+ ]
+ }
+ ],
+ "prompt_number": 27
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 10.6 Page no 560"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "d=6.4*10**6 #m\n",
+ "B=0.4*10**-4 #T\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "M=(B*4*math.pi*d**3)/u\n",
+ "\n",
+ "#Result\n",
+ "print\"Earth's dipole moment is\", round(M*10**-23,2)*10**23,\"Am**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Earth's dipole moment is 1.05e+23 Am**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 33
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 10.7 Page no 560"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "M=0.40\n",
+ "d=0.5\n",
+ "\n",
+ "#Calculation\n",
+ "Beqa=u*M/(4*math.pi*d**3)\n",
+ "Baxial=2*Beqa\n",
+ "\n",
+ "#Result\n",
+ "print\"Magnitude of axial field is\", Baxial,\"T\"\n",
+ "print\"Magnitude of equatorial field is\",Beqa,\"T\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Magnitude of axial field is 6.4e-07 T\n",
+ "Magnitude of equatorial field is 3.2e-07 T\n"
+ ]
+ }
+ ],
+ "prompt_number": 39
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 10.8 Page no 560"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "e=1.6*10**-19\n",
+ "f=6.8*10**15\n",
+ "n=1\n",
+ "r=0.53*10**-10\n",
+ "\n",
+ "#Calculation \n",
+ "import math\n",
+ "I=e*f\n",
+ "M=n*I*math.pi*r**2\n",
+ "\n",
+ "#Result\n",
+ "print\"Equivalent magnetic moment is\", round(M*10**24,1)*10**-24,\"Am**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Equivalent magnetic moment is 9.6e-24 Am**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 46
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 10.9 Page no 561"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "n=50\n",
+ "r=0.2 #m\n",
+ "I=12 #A\n",
+ "\n",
+ "#Calculation\n",
+ "B=(u*n*I)/(2.0*r)\n",
+ "M=n*I*math.pi*r**2\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Magnetic field at the centre of the coil is\", round(B*10**3,3),\"*10**-3 T\"\n",
+ "print\"(ii) Magnetic moment is\",round(M,1),\"Am**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Magnetic field at the centre of the coil is 1.885 *10**-3 T\n",
+ "(ii) Magnetic moment is 75.4 Am**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 58
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 10.10 Page no 561"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "A=7.5*10**-4 #m**2\n",
+ "I=12 #A\n",
+ "\n",
+ "#Calculation\n",
+ "M=A*I\n",
+ "\n",
+ "#Result\n",
+ "print\"Magnitude of the magnetic moment is\", M*10**3,\"*10**-3 Am**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Magnitude of the magnetic moment is 9.0 *10**-3 Am**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 62
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 10.11 Page no 561"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "n=100\n",
+ "I=0.1 #A\n",
+ "r=0.05\n",
+ "B=1.5 #T\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "M=n*I*math.pi*r**2\n",
+ "W=2*M*B\n",
+ "\n",
+ "#Result\n",
+ "print\"Magnitude of the coil is\", round(M,4),\"Am**2\"\n",
+ "print\"Workdone is\",round(W,4),\"J\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Magnitude of the coil is 0.0785 Am**2\n",
+ "Workdone is 0.2356 J\n"
+ ]
+ }
+ ],
+ "prompt_number": 70
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 10.12 Page no 561"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "n=10\n",
+ "I=3\n",
+ "A=7.85*10**-3\n",
+ "B=10**-2 #T\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "M=n*I*A\n",
+ "U1=-M*B*math.cos(0)\n",
+ "Uf=-M*B*math.cos(90)\n",
+ "w=-U1\n",
+ "t=M*B*math.sin(90*3.14/180.0)\n",
+ "\n",
+ "#Result\n",
+ "print\"Work done is\", round(t*10**3,1),\"*10**-3 Nm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Work done is 2.4 *10**-3 Nm\n"
+ ]
+ }
+ ],
+ "prompt_number": 81
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 10.13 Page no 562"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "M=4.8*10**-2 #J/T\n",
+ "a=30 #degree\n",
+ "B=3*10**-2 #t\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "t=M*B*math.sin(a*3.14/180.0)\n",
+ "\n",
+ "#Result\n",
+ "print\"Torque acting on the needle is\", round(t*10**4,1),\"*10**-4 Nm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Torque acting on the needle is 7.2 *10**-4 Nm\n"
+ ]
+ }
+ ],
+ "prompt_number": 86
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 10.14 Page no 562"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "B=0.2 #T\n",
+ "a=30 #degree\n",
+ "t=0.06 #Nm\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "M=t/(B*math.sin(a*3.14/180.0))\n",
+ "U=M*B*math.cos(1*3.14/180.0)\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Magnetic moment of the magnet is\", round(M,1),\"Am**2\"\n",
+ "print\"(ii) Orientation of the magnet is\", round(U,0)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Magnetic moment of the magnet is 0.6 Am**2\n",
+ "(ii) Orientation of the magnet is 0.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 97
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 10.15 Page no 562"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#given\n",
+ "a=30 #degree\n",
+ "B=800*10**-4 #T\n",
+ "t=0.016 #Nm\n",
+ "A=2*10**-4 #m**2\n",
+ "n=1000 #turns\n",
+ "\n",
+ "#Calculation\n",
+ "M=t/(B*math.sin(a*3.14/180.0))\n",
+ "W=2*M*B\n",
+ "I=M/(n*A)\n",
+ "\n",
+ "#Result\n",
+ "print\"(a) Magnetic moment of the magnet is\", round(M,2),\"Am**2\"\n",
+ "print\"(b) Work done is\", round(W,3),\"J\"\n",
+ "print\"(c) Current flowing through the solenoid is\", round(I,0),\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) Magnetic moment of the magnet is 0.4 Am**2\n",
+ "(b) Work done is 0.064 J\n",
+ "(c) Current flowing through the solenoid is 2.0 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 108
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 10.16 Page no 563"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "t=6.70\n",
+ "n=10.0\n",
+ "I=7.5*10**-6 #Kgm**2\n",
+ "M=6.7*10**-2 #Am**2\n",
+ "\n",
+ "#Calculation\n",
+ "T=t/n\n",
+ "B=(4*math.pi**2*I)/(M*T**2)\n",
+ "\n",
+ "#Result\n",
+ "print\"Magnitude of the magnetic field is\", round(B,2),\"T\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Magnitude of the magnetic field is 0.01 T\n"
+ ]
+ }
+ ],
+ "prompt_number": 113
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 10.18 Page no 569"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "t=1.2*10**-3 #nm\n",
+ "M=60\n",
+ "H=40*10**-6\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "A=t/(M*H)\n",
+ "a=math.asin(A)*180/3.14\n",
+ "\n",
+ "#Result\n",
+ "print\"Angle of the declination is\", round(a,0),\"degree\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Angle of the declination is 30.0 degree\n"
+ ]
+ }
+ ],
+ "prompt_number": 126
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 10.19 Page no 569"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "V=math.sqrt(3)\n",
+ "\n",
+ "#calculation\n",
+ "import math\n",
+ "a=math.atan(V)*180/3.14\n",
+ "\n",
+ "#Result\n",
+ "print\"Angle of dip is\", round(a,0),\"Degree\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Angle of dip is 60.0 Degree\n"
+ ]
+ }
+ ],
+ "prompt_number": 131
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 10.20 Page no 569"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "H=0.28 #G\n",
+ "V=0.40 #G\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "A=V/H\n",
+ "a=math.atan(A)*180/3.14\n",
+ "R=math.sqrt(H**2+V**2)\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Angle of dip is\", round(a,0),\"Degree\"\n",
+ "print\"(ii) Earth's total magnetic field is\", round(R,2),\"G\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Angle of dip is 55.0 Degree\n",
+ "(ii) Earth's total magnetic field is 0.49 G\n"
+ ]
+ }
+ ],
+ "prompt_number": 140
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 10.22 Page no 570"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "H=0.40\n",
+ "a=18 #degree\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "R=H/(math.cos(a*3.14/180.0))\n",
+ "\n",
+ "#Result\n",
+ "print\"Magnitude of earth's magnetic field is\", round(R,2),\"G\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Magnitude of earth's magnetic field is 0.42 G\n"
+ ]
+ }
+ ],
+ "prompt_number": 145
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 10.24 Page no 571"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=45 #Degree\n",
+ "b=60 #Degree\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "A=math.tan(a*3.14/180.0)/(math.cos(b*3.14/180.0))\n",
+ "a=math.atan(A)*180/3.14\n",
+ "\n",
+ "#Result\n",
+ "print\"Apparant dip is\", round(a,1),\"Degree\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Apparant dip is 63.4 Degree\n"
+ ]
+ }
+ ],
+ "prompt_number": 152
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 10.25 Page no 574"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "M=1.6 #Am**2\n",
+ "d=0.20 #m\n",
+ "u=10**-7 #N/A**2\n",
+ "\n",
+ "#Calculation\n",
+ "H=u*2*M/(d**3)\n",
+ "\n",
+ "#Result\n",
+ "print\"Horizontal component of the earth's magnetic field is\", H,\"T\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Horizontal component of the earth's magnetic field is 4e-05 T\n"
+ ]
+ }
+ ],
+ "prompt_number": 155
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 10.26 Page no 574"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "l=0.05 #m\n",
+ "d=0.12 #m\n",
+ "H=0.34*10**-4 #T\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "u=4*math.pi*10**-7\n",
+ "M=(4*math.pi*H*(d**2+l**2)**1.5)/u\n",
+ "\n",
+ "#Result\n",
+ "print\"Magnetic moment of the magnet is\", round(M,3),\"J/T\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Magnetic moment of the magnet is 0.747 J/T\n"
+ ]
+ }
+ ],
+ "prompt_number": 162
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 10.27 Page no 577"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "r=7*10**-2 #m\n",
+ "H=2*10**-5 #T\n",
+ "n=50\n",
+ "\n",
+ "#calculation\n",
+ "import math\n",
+ "l=(2*r*H*math.tan(45*180/3.14))/u*n\n",
+ "\n",
+ "#Result\n",
+ "print\"Value of current is\", round(l*10**-3,3),\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Value of current is 0.043 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 172
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 10.28 Page no 577"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "K=0.095 #A\n",
+ "n=50\n",
+ "r=10*10**-2 #m\n",
+ "\n",
+ "#Calculation\n",
+ "H=K*u*n/(2.0*r)\n",
+ "\n",
+ "#Result\n",
+ "print\"Horizontal component of earth's magnetic field is\", round(H*10**4,3),\"*10**-4 T\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Horizontal component of earth's magnetic field is 0.298 *10**-4 T\n"
+ ]
+ }
+ ],
+ "prompt_number": 178
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 10.30 Page no 577"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=30 #degree\n",
+ "b=45 #degree\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "m=(2*math.tan(a*3.14/180.0))/(math.tan(b*3.14/180.0))\n",
+ "\n",
+ "#Result\n",
+ "print\"Ratio of number of turns of the tangent galvanometers\", round(m,3)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Ratio of number of turns of the tangent galvanometers 1.155\n"
+ ]
+ }
+ ],
+ "prompt_number": 188
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/Chapter28_1.ipynb b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/Chapter28_1.ipynb new file mode 100644 index 00000000..afb57de5 --- /dev/null +++ b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/Chapter28_1.ipynb @@ -0,0 +1,59 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:511d2d405e1ede92783e0ff6e1c085ebc325e49ab2eff49fe438f3081d70cb4d"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter28 Digital Electronics"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "Example 28.3 page no 1497"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": []
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=1\n",
+ "\n",
+ "#Calculation\n",
+ "A=a*2**5+a*2**4+a*2**0\n",
+ "\n",
+ "#Result\n",
+ "print\"equivilant decimal is\",A"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "equivilant decimal is 49\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/Chapter29_1.ipynb b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/Chapter29_1.ipynb new file mode 100644 index 00000000..9f1a9134 --- /dev/null +++ b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/Chapter29_1.ipynb @@ -0,0 +1,526 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:8b87017ad47964520c2ead85c01092623a6e1bffcb1688b462701d086beba4f8"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter29 Communication System"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 29.1 page no 1550"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "c=3*10**8\n",
+ "f=30.0*10**6\n",
+ "f1=300*10**6\n",
+ "f2=3000*10**6\n",
+ "\n",
+ "#Calculation\n",
+ "l=c/f\n",
+ "l1=l/2.0\n",
+ "l2=c/f1\n",
+ "l3=l2/2.0\n",
+ "l4=c/f2\n",
+ "l5=l4/2.0\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) length of half wave dipole antenna at 30 MHz is\",l1,\"m\"\n",
+ "print\"(ii) length of half wave dipole antenna at 300 MHz is\",l3,\"m\"\n",
+ "print\"(iii) length of half wave dipole antenna at 3000 MHz is\",15,\"m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) length of half wave dipole antenna at 30 MHz is 5.0 m\n",
+ "(ii) length of half wave dipole antenna at 300 MHz is 0.5 m\n",
+ "(iii) length of half wave dipole antenna at 3000 MHz is 15 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 29.2 page no 1550"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "c=3*10**8 #m/s\n",
+ "f=3*10**4\n",
+ "f1=6*10**6\n",
+ "f2=5*10**7\n",
+ "\n",
+ "#Calculation\n",
+ "l=c/(4.0*f)\n",
+ "l1=c/(4.0*f1)\n",
+ "l2=c/(4.0*f2)\n",
+ "\n",
+ "#Result \n",
+ "print\"(i) length of quarter wave dipole antenna at 3*10**4 is\",l,\"m\"\n",
+ "print\"(ii) length of quarter wave dipole antenna at 6*10**6 is\",l1,\"m\"\n",
+ "print\"(iii) length of quarter wave dipole antena at 5*10**7 is\",l2,\"m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) length of quarter wave dipole antenna at 3*10**4 is 2500.0 m\n",
+ "(ii) length of quarter wave dipole antenna at 6*10**6 is 12.5 m\n",
+ "(iii) length of quarter wave dipole antena at 5*10**7 is 1.5 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 29.3 page no 1551"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "AM=16 #mV\n",
+ "AM1=4 #mV\n",
+ "\n",
+ "#Calculation\n",
+ "Vmax=AM/2.0\n",
+ "Vmin=AM1/2.0\n",
+ "Ma=(Vmax-Vmin)/(Vmax+Vmin)\n",
+ "\n",
+ "#Result\n",
+ "print\" The modulation factor is\",Ma"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " The modulation factor is 0.6\n"
+ ]
+ }
+ ],
+ "prompt_number": 25
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 29.4 page no 1551"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Es=50 #V\n",
+ "Ec=100.0 #V\n",
+ "\n",
+ "#Calculation\n",
+ "Ma=Es/Ec\n",
+ "\n",
+ "#Result\n",
+ "print\"The modulation factor\",Ma"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The modulation factor 0.5\n"
+ ]
+ }
+ ],
+ "prompt_number": 28
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 29.6 page no 1552"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Pc=500 #watts\n",
+ "\n",
+ "#Calculation\n",
+ "Ps=(1/2.0)*(Pc)\n",
+ "Pt=Pc+Ps\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) sideband power is\",Ps,\"W\"\n",
+ "print\"(ii) power of AM wave is\",Pt,\"W\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) sideband power is 250.0 W\n",
+ "(ii) power of AM wave is 750.0 W\n"
+ ]
+ }
+ ],
+ "prompt_number": 36
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 29.7 page no 1552"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Pc=50\n",
+ "Ma=0.8\n",
+ "Ma1=0.1\n",
+ "\n",
+ "#Calculation\n",
+ "Ps=(1/2.0)*Ma**2*Pc\n",
+ "Ps1=(1/2.0)*Ma1**2*Pc\n",
+ "\n",
+ "#Result\n",
+ "print\"total sideband at 80% is\",Ps,\"KW\"\n",
+ "print\"total sideband at 10% is\",Ps1,\"KW\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "total sideband at 80% is 16.0 KW\n",
+ "total sideband at 10% is 0.25 KW\n"
+ ]
+ }
+ ],
+ "prompt_number": 42
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 29.8 page no 1552"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Fc=500 #KHz\n",
+ "Fs=1 #KHz\n",
+ "\n",
+ "#Calculation\n",
+ "A1=Fc+Fs\n",
+ "A2=Fc-Fs\n",
+ "B=A1-A2\n",
+ "\n",
+ "#Result\n",
+ "print\"sideband frequancies are\",A1,\"KHz and\",A2,\"KHz\"\n",
+ "print\"bandwidth required is\",B,\"KHz\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "sideband frequancies are 501 KHz and 499 KHz\n",
+ "bandwidth required is 2 KHz\n"
+ ]
+ }
+ ],
+ "prompt_number": 47
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 29.9 page no 1552"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "F=10**10 #Hz\n",
+ "D=8*10**3 #Hz\n",
+ "\n",
+ "#Calculation\n",
+ "B=2/100.0*10**10\n",
+ "C=B/D\n",
+ "\n",
+ "#Result\n",
+ "print\"No. of telephones channels are\",C*10**-4,\"10**4\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "No. of telephones channels are 2.5 10**4\n"
+ ]
+ }
+ ],
+ "prompt_number": 56
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 20.10 page no 1553"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "L=800.0*10**-7\n",
+ "C=3.0*10**8\n",
+ "f1=4.5*10**6 #Hz\n",
+ "\n",
+ "#Calculation\n",
+ "f=C/L\n",
+ "d=(1/100.0)*f\n",
+ "E=d/L\n",
+ "G=d/f1\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) number of channels for audio signal is\",round(E*10**-14,1),\"*10**8\"\n",
+ "print\"(ii) number of channels for video tv signal is\",round(G*10**-3,1),\"*10**5\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) number of channels for audio signal is 4.7 *10**8\n",
+ "(ii) number of channels for video tv signal is 8.3 *10**5\n"
+ ]
+ }
+ ],
+ "prompt_number": 80
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 20.11 Page no 1565"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R=6400*10**3 #m\n",
+ "h=160\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "d=math.sqrt(2*R*h)\n",
+ "h2=4*h\n",
+ "\n",
+ "#Result\n",
+ "print\"Height is\", h2,\"m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Height is 640 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 85
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 20.12 Page no 1565"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R=6.4*10**6 #m\n",
+ "h=110\n",
+ "\n",
+ "#Calculation \n",
+ "import math\n",
+ "d=(math.sqrt(2*R*h))*10**-3\n",
+ "P=math.pi*d**2\n",
+ "\n",
+ "#Result\n",
+ "print\"Population covered is\", round(P*10**-3,1),\"*10**6\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Population covered is 4.4 *10**6\n"
+ ]
+ }
+ ],
+ "prompt_number": 93
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 20.13 Page no 1565"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R=6.4*10**6 #m\n",
+ "hr=50 #m\n",
+ "ht=32 #m\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "d=math.sqrt(2*R*ht)+math.sqrt(2*R*hr)\n",
+ "\n",
+ "#Result\n",
+ "print\"Maximum distance is\", round(d*10**-3,1),\"Km\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Maximum distance is 45.5 Km\n"
+ ]
+ }
+ ],
+ "prompt_number": 99
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 29.14 Page no 1566"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "h=300\n",
+ "R=6.4*10**6 #m\n",
+ "N=10**12\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "d=math.sqrt(2*R*h)\n",
+ "fc=9*N**0.5\n",
+ "\n",
+ "#Result\n",
+ "print\"fc=\", fc*10**-6,\"MHz\"\n",
+ "print\"5 MHz comes via ionospheric propogation.and 100 MHz signal comes via satellite transmission.\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "fc= 9.0 MHz\n",
+ "5 MHz comes via ionospheric propogation.and 100 MHz signal comes via satellite transmission.\n"
+ ]
+ }
+ ],
+ "prompt_number": 107
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter11_1.ipynb b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter11_1.ipynb new file mode 100644 index 00000000..11322719 --- /dev/null +++ b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter11_1.ipynb @@ -0,0 +1,467 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:f9cf5cb53006209575af5d70cc318cffdaba99568e9075844fb3e2f1810bf06f"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 11 Classification of magnetic materials"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 11.1 Page no 614"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "e=1.6*10**-19\n",
+ "f=6.8*10**15\n",
+ "n=1\n",
+ "r=0.53*10**-10\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "I=e*f\n",
+ "M=n*I*math.pi*r**2\n",
+ "\n",
+ "#Result\n",
+ "print\"Equivalent magnetic moment is\", round(M*10**24,1)*10**-24,\"Am**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Equivalent magnetic moment is 9.6e-24 Am**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 11.2 Page no 615"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "E=240\n",
+ "R=474.0\n",
+ "r=12.5*10**-2\n",
+ "N=500\n",
+ "ur=5000\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "I=E/R\n",
+ "I1=2*math.pi*r\n",
+ "H=(N*I)/I1\n",
+ "u=4*math.pi*10**-7\n",
+ "B=u*ur*H\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) The magnetising force is\", round(H,0),\"AT/m\"\n",
+ "print\"(ii) The magnetic flux density is\",round(B,2),\"Wb/m**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) The magnetising force is 322.0 AT/m\n",
+ "(ii) The magnetic flux density is 2.03 Wb/m**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 18
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 11.3 Page no 615"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "r1=11\n",
+ "r2=12\n",
+ "B=2.5 #T\n",
+ "a=3000\n",
+ "I=0.70 #A\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "r=((r1+r2)/2.0)*10**-2\n",
+ "n=a/(2*math.pi*r)\n",
+ "ur=B*2*math.pi*r/(4*math.pi*10**-7*a*I)\n",
+ "\n",
+ "#Result\n",
+ "print\"Relative permeability is\", round(ur,1)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Relative permeability is 684.5\n"
+ ]
+ }
+ ],
+ "prompt_number": 24
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 11.4 Page no 616"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "I=0.5 #m\n",
+ "N=500\n",
+ "I1=0.15 #A\n",
+ "a=5000\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "H=(N*I1)/I\n",
+ "B=4*math.pi*10**-7*H\n",
+ "B1=B*a\n",
+ "I3=(B1-(H*4*math.pi*10**-7))/(4.0*math.pi*10**-7)\n",
+ "\n",
+ "#Result\n",
+ "print round(I3*10**-5,1),\"*10**5 A/m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "7.5 *10**5 A/m\n"
+ ]
+ }
+ ],
+ "prompt_number": 35
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 11.5 Page no 616"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "B=0.6\n",
+ "H=360.0\n",
+ "\n",
+ "#Calculation\n",
+ "u=B/H\n",
+ "x=(u-1*4*math.pi*10**-7)/(4.0*math.pi*10**-7)\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Permeability is\",round(u*10**3,2),\"*10**-3 T/A m\"\n",
+ "print\"(ii) Susceptibility of the material is\",round(x,0)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Permeability is 1.67 *10**-3 T/A m\n",
+ "(ii) Susceptibility of the material is 1325.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 46
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 11.6 Page no 616"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "M=8.0*10**22 #Am**2\n",
+ "R=64*10**5 #m\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "I=(3*M)/(4.0*math.pi*R**3)\n",
+ "\n",
+ "#Result\n",
+ "print\"Earth's magnetisation is\", round(I,1),\"A/m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Earth's magnetisation is 72.9 A/m\n"
+ ]
+ }
+ ],
+ "prompt_number": 52
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 11.7 Page no 616"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#given\n",
+ "N=1800\n",
+ "l=0.6\n",
+ "I=0.9 #A\n",
+ "ur=500\n",
+ "n1=6.02*10**26\n",
+ "a=55.85\n",
+ "y=7850\n",
+ "\n",
+ "#Calculation\n",
+ "n=N/l\n",
+ "H=n*I\n",
+ "I1=(ur-1)*H\n",
+ "B=4*math.pi*10**-7*ur*H\n",
+ "x=(y*n1)/a\n",
+ "X=I1/x\n",
+ "\n",
+ "#Result\n",
+ "print\"Average magnetic moment per iron atom is\", round(X*10**23,2)*10**-23,\" A m**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Average magnetic moment per iron atom is 1.59e-23 A m**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 63
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 11.8 Page no 617"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "M=8.4 #g\n",
+ "d=7200.0\n",
+ "f=50 #Hz\n",
+ "E=3.2*10**4\n",
+ "t=30*60.0\n",
+ "\n",
+ "#Calculation\n",
+ "V=M/d\n",
+ "P=E/t\n",
+ "E1=P/(V*f)\n",
+ "\n",
+ "#Result\n",
+ "print\"Energy dissipated per unit volume is\", round(E1,0),\"J/m**3/cycle\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Energy dissipated per unit volume is 305.0 J/m**3/cycle\n"
+ ]
+ }
+ ],
+ "prompt_number": 70
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 11.9 Page no 617"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "H=4*10**3 #A/m\n",
+ "a=60\n",
+ "b=0.12\n",
+ "\n",
+ "#Calculation\n",
+ "n=a/b\n",
+ "I=H/n\n",
+ "\n",
+ "#Result\n",
+ "print\"Current should be sent through the solenoid is\", I,\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Current should be sent through the solenoid is 8.0 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 73
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 11.10 Page no 617"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "x=1.68*10**-4\n",
+ "T1=293\n",
+ "T2=77.4\n",
+ "\n",
+ "#Calculation\n",
+ "x1=(x*T1)/T2\n",
+ "\n",
+ "#Result\n",
+ "print\"Susceptibility is\", round(x1*10**4,2),\"*10**-4\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Susceptibility is 6.36 *10**-4\n"
+ ]
+ }
+ ],
+ "prompt_number": 79
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 11.11 Page no 617"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "l=10**-6 #m\n",
+ "d=7.9 #g\n",
+ "a=6.023*10**23\n",
+ "n=55.0\n",
+ "M1=9.27*10**-24\n",
+ "\n",
+ "#Calculation\n",
+ "V=l**2\n",
+ "M=V*d\n",
+ "N=(a*M)/n\n",
+ "Mmax=N*M1\n",
+ "Imax=Mmax/V*10**-4\n",
+ "\n",
+ "#Result\n",
+ "print\"Number of iron atom is\",round(N*10**-10,2),\"*10**10 atoms\"\n",
+ "print\"Magnetisation of the dipole is\",round(Imax*10**5,0),\"*10**5 A/m\"\n",
+ "print\"Maximum possible dipole moment is\",round(Mmax*10**13,0)*10**-13,\"A m**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Number of iron atom is 8.65 *10**10 atoms\n",
+ "Magnetisation of the dipole is 8.0 *10**5 A/m\n",
+ "Maximum possible dipole moment is 8e-13 A m**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 113
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter12_1.ipynb b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter12_1.ipynb new file mode 100644 index 00000000..c7cfe98a --- /dev/null +++ b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter12_1.ipynb @@ -0,0 +1,1049 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:58b6e0817b83a6d5c266f15dbe5c66e891c1c2ed8d1ef60dafaf43f98ff2d620"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 12 Electromagnetic induction"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 12.1 Page no 665"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=20 #mWb\n",
+ "a1=-20 #mWb\n",
+ "t=2*10**-3 #s\n",
+ "N=100\n",
+ "\n",
+ "#Calculation\n",
+ "a2=(a1-a)*10**-3\n",
+ "e=(-N*a2)/t\n",
+ "\n",
+ "#Result\n",
+ "print\"Average e.m.f induced in the coil is\", e,\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Average e.m.f induced in the coil is 2000.0 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 12.2 Page no 665"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "r=5*10**-2 #m\n",
+ "N=1\n",
+ "B=0.35\n",
+ "t=0.12 #S\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "A=math.pi*r**2\n",
+ "a1=B*A\n",
+ "a2=-B*a1\n",
+ "e=(N*a1)/t\n",
+ "\n",
+ "#Result\n",
+ "print round(e,2),\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "0.02 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 12
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 12.3 Page no 665"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "A=10**-2 #m**2\n",
+ "a=45 #degree\n",
+ "B1=0.1 #T\n",
+ "R=0.5 #ohm\n",
+ "t=0.7 #S\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "a1=B1*A*math.cos(a*3.14/180.0)\n",
+ "a2=0\n",
+ "a3=a1-a2\n",
+ "e=a3/t\n",
+ "I=e/R\n",
+ "\n",
+ "#Result\n",
+ "print\"Current during this time interval is\", round(I*10**3,1),\"*10**-3 A\"\n",
+ "print\"Magnitude of induced emf is\",round(e*10**3,0),\"*10**-3 V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Current during this time interval is 2.0 *10**-3 A\n",
+ "Magnitude of induced emf is 1.0 *10**-3 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 24
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 12.4 Page no 666"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "I=1.2*10**-3 #A\n",
+ "N=1.0\n",
+ "R=10 #ohm\n",
+ "\n",
+ "#Calculation\n",
+ "e=I*R\n",
+ "a=e/N\n",
+ "\n",
+ "#Result\n",
+ "print\"Necessary rate is\", a*10**2,\"*10**-2 Wb/second\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Necessary rate is 1.2 *10**-2 Wb/second\n"
+ ]
+ }
+ ],
+ "prompt_number": 28
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 12.5 Page no 666"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "r=10**-1 #m\n",
+ "B=3.0*10**-5 #T\n",
+ "t=0.25 #S\n",
+ "N=500\n",
+ "R=2 #ohm\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "a1=B*math.pi*r**2*math.cos(0*3.14/180.0)\n",
+ "a2=B*math.pi*r**2*math.cos(180*3.14/180.0)\n",
+ "a3=a1-a2\n",
+ "e=(N*a3)/t\n",
+ "I=e/R\n",
+ "\n",
+ "#Result\n",
+ "print\"Magnitude of the emf is\", round(e*10**3,1),\"*10**-3 V\"\n",
+ "print\"Current induced in the coil is\",round(I*10**3,1),\"*1)**-3 A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Magnitude of the emf is 3.8 *10**-3 V\n",
+ "Current induced in the coil is 1.9 *1)**-3 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 47
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 12.6 Page no 666"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "e=10**-2 #V\n",
+ "B=5*10**-5 #T\n",
+ "r=0.5 #m\n",
+ "N=1\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "A=math.pi*r**2\n",
+ "n=(e*N)/(math.pi*r**2*B)\n",
+ "\n",
+ "#Result\n",
+ "print\"Rate of rotation of the blade is\", round(n,1),\"revolutions/second\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Rate of rotation of the blade is 254.6 revolutions/second\n"
+ ]
+ }
+ ],
+ "prompt_number": 53
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 12.7 Page no 667"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=12\n",
+ "b=7\n",
+ "t=2\n",
+ "\n",
+ "#Calculation\n",
+ "e=((a*t)+b)*10**-3\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Magnitude of induced emf is\", e*10**3,\"mV\"\n",
+ "print\"(ii) The current induced in the coil will be anticlockwise\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Magnitude of induced emf is 31.0 mV\n",
+ "(ii) The current induced in the coil will be anticlockwise\n"
+ ]
+ }
+ ],
+ "prompt_number": 59
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 12.8 Page no 673"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "B=1 #T\n",
+ "l=0.5 #m\n",
+ "v=40 #m/s\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "e=B*l*v*math.sin(60*3.14/180.0)\n",
+ "\n",
+ "#Result\n",
+ "print\"emf induced in the conductor is\", round(e,2)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "emf induced in the conductor is 17.32\n"
+ ]
+ }
+ ],
+ "prompt_number": 63
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 12.9 Page no 673"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "g=9.8\n",
+ "h=10\n",
+ "B=1.7*10**-5\n",
+ "l=1 #m\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "v=math.sqrt(2*g*h)\n",
+ "e=B*l*v\n",
+ "\n",
+ "#Result\n",
+ "print\"Potential difference between its end is\", e*10**4,\"*10**4 V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Potential difference between its end is 2.38 *10**4 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 68
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 12.10 Page no 673"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "v=72 *(5/18.0) #Km/h\n",
+ "B=40*10**-6 #T\n",
+ "A=40\n",
+ "l=2 #m\n",
+ "t=1.0\n",
+ "N=1\n",
+ "\n",
+ "#Calculation\n",
+ "A=l*v\n",
+ "a=B*A\n",
+ "e=N*a/t\n",
+ "\n",
+ "#Result\n",
+ "print\"e.m.f generated in the axle of the car\", e*10**3,\"mV\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "e.m.f generated in the axle of the car 1.6 mV\n"
+ ]
+ }
+ ],
+ "prompt_number": 77
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 12.11 Page no 673"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "w=1000/60.0\n",
+ "r=0.3\n",
+ "B=0.5 #T\n",
+ "\n",
+ "#Calculation\n",
+ "v=w*r\n",
+ "vav=v/2.0\n",
+ "e=B*r*vav\n",
+ "\n",
+ "#Result\n",
+ "print\"e.m.f induced is\",e,\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "e.m.f induced is 0.375 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 84
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 12.12 Page no "
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "r=0.5 #m\n",
+ "n=2 #r.p.s\n",
+ "B=0.4*10**-4 #T\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "w=2*math.pi*n\n",
+ "e=0.5*B*r**2*w\n",
+ "\n",
+ "#Result\n",
+ "print\"Magnitude of induced e.m.f between the axle and rim is\", round(e*10**5,2)*10**-5,\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Magnitude of induced e.m.f between the axle and rim is 6.28e-05 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 91
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 12.13 Page no 674"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R=1 #m\n",
+ "B=1\n",
+ "f=50\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "e=math.pi*R**2*B*f\n",
+ "\n",
+ "#Result\n",
+ "print\"e.m.f between the centre and the matallic ring is\", round(e,1),\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "e.m.f between the centre and the matallic ring is 157.1 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 96
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 12.14 Page no 679"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "N=500\n",
+ "a=1.4*10**-4 #Wb\n",
+ "l=2.5 #A\n",
+ "\n",
+ "#Calculation\n",
+ "L=(N*a)/l\n",
+ "\n",
+ "#Result\n",
+ "print\"Inductance of the coil is\", L*10**3,\"mH\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Inductance of the coil is 28.0 mH\n"
+ ]
+ }
+ ],
+ "prompt_number": 100
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 12.15 Page no 679"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "L=130*10**-3 #H\n",
+ "I1=20 #mA\n",
+ "I2=28 #mA\n",
+ "t=140.0*10**-3 #S\n",
+ "\n",
+ "#Calculation\n",
+ "l=I2-I1\n",
+ "e=(-L*l)/t\n",
+ "\n",
+ "#Result\n",
+ "print\"Magnitude of induced e.m.f is\", round(e,2),\"*10**-3 V\"\n",
+ "print\"Direction oppose the increase in current\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Magnitude of induced e.m.f is -7.43 *10**-3 V\n",
+ "Direction oppose the increase in current\n"
+ ]
+ }
+ ],
+ "prompt_number": 106
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 12.16 Page no 679"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "N=4000\n",
+ "l=0.6 #m\n",
+ "r=16*10**-4 #m\n",
+ "\n",
+ "#Calculation\n",
+ "u=4*math.pi*10**-7\n",
+ "L=(u*N**2*((math.pi*r)/4.0))/l\n",
+ "Liron=N*L\n",
+ "\n",
+ "#Result\n",
+ "print\"Inductance of the solenoid is\", round(Liron,0),\"H\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Inductance of the solenoid is 168.0 H\n"
+ ]
+ }
+ ],
+ "prompt_number": 115
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 12.17 Page no 679"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "L=10.0 #H\n",
+ "e=300 #V\n",
+ "t=10**-2 #S\n",
+ "\n",
+ "#Calculation\n",
+ "dl=(e*t)/L\n",
+ "a=e*t\n",
+ "\n",
+ "#Result\n",
+ "print\"Charge in magnetic flux is\", a,\"Wb\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Charge in magnetic flux is 3.0 Wb\n"
+ ]
+ }
+ ],
+ "prompt_number": 119
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 12.18 Page no 680"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "L=10*10**-3\n",
+ "I=4*10**-3\n",
+ "N=200.0\n",
+ "\n",
+ "#Calculation\n",
+ "N1=L*I\n",
+ "a=N1/N\n",
+ "\n",
+ "#Result\n",
+ "print\"Total flux linked with the coil is\", N1,\"Wb\"\n",
+ "print\"Magnetic flux through the cross section of the coil is\",a,\"Wb\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Total flux linked with the coil is 4e-05 Wb\n",
+ "Magnetic flux through the cross section of the coil is 2e-07 Wb\n"
+ ]
+ }
+ ],
+ "prompt_number": 125
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 12.19 Page no 680"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "L=500*10**-3\n",
+ "I1=20*10**-3 #A\n",
+ "I2=10*10**-3 #A\n",
+ "\n",
+ "#Calculation\n",
+ "U1=0.5*L*I1**2\n",
+ "U2=0.5*L*I2**2\n",
+ "\n",
+ "#Result\n",
+ "print \"Magnetic energy stored in the coil is\",U1*10**6,\"*10**-4 J\"\n",
+ "print\"New value of energy is\",U2,\"J\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Magnetic energy stored in the coil is 100.0 *10**-4 J\n",
+ "New value of energy is 2.5e-05 J\n"
+ ]
+ }
+ ],
+ "prompt_number": 134
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 12.20 Page no 680"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "E=12\n",
+ "R=30.0 #ohm\n",
+ "L=0.22 \n",
+ "\n",
+ "#Calculation\n",
+ "I0=E/R\n",
+ "I=I0/2.0\n",
+ "P=E*I\n",
+ "dl=(E-(I*R))/L\n",
+ "du=L*I*dl\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Energy being delivered by the battery is\", P,\"W\"\n",
+ "print\"(ii) ENergy being stored in the magnetic field of inductor is\",du,\"W\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Energy being delivered by the battery is 2.4 W\n",
+ "(ii) ENergy being stored in the magnetic field of inductor is 1.2 W\n"
+ ]
+ }
+ ],
+ "prompt_number": 144
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 12.21 Page no 680"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "L=2.0 #H\n",
+ "i=2 #A\n",
+ "\n",
+ "#Calculation\n",
+ "U=0.5*L*i**2\n",
+ "\n",
+ "#Result\n",
+ "print\"Amount of energy spent during the period is\", U,\"J\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Amount of energy spent during the period is 4.0 J\n"
+ ]
+ }
+ ],
+ "prompt_number": 147
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 12.22 Page no 686"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "e=1500 #V\n",
+ "dl=3 #A\n",
+ "dt=0.001 #s\n",
+ "\n",
+ "#Calculation\n",
+ "M=(e*dt)/dl\n",
+ "\n",
+ "#Result\n",
+ "print\"Mumtual induction between the two coils is\", M,\"H\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Mumtual induction between the two coils is 0.5 H\n"
+ ]
+ }
+ ],
+ "prompt_number": 150
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 12.23 Page no 686"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "N2=1000\n",
+ "I1=5.0 #A\n",
+ "a2=0.4*10**-4 #Wb\n",
+ "dl=-24 #A\n",
+ "dt=0.02 #S\n",
+ "\n",
+ "#Calculation\n",
+ "M=(N2*a2)/I1\n",
+ "eb=(-M*dl)/dt\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Mutual induction between A and B is\", M,\"H\"\n",
+ "print\"(ii) e.m.f induced by the coil is\", eb"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Mutual induction between A and B is 0.008 H\n",
+ "(ii) e.m.f induced by the coil is 9.6\n"
+ ]
+ }
+ ],
+ "prompt_number": 155
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 12.24 Page no 687"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "N=1200\n",
+ "A=12*10**-4 #m**2\n",
+ "r=0.15 #m\n",
+ "N2=300\n",
+ "a=0.05\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "u=4*math.pi*10**-7\n",
+ "L=(u*N**2*A)/(2*math.pi*r)\n",
+ "M=(u*N*N2*A)/(2*math.pi*r)\n",
+ "dl=2/a\n",
+ "e=M*dl\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Self inductance of the toroid is\", round(L*10**3,1),\"*10**-3 H\"\n",
+ "print\"(ii) Induced e.m.f. in the second coil is\",round(e,3),\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Self inductance of the toroid is 2.3 *10**-3 H\n",
+ "(ii) Induced e.m.f. in the second coil is 0.023 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 168
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 12.25 Page no 688"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "I=2.0\n",
+ "a1=20*10**-2\n",
+ "x=0.15\n",
+ "A2=0.3*10**-2\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "u=4*math.pi*10**-7\n",
+ "B1=(u*I*a1**2)/(2.0*(a1**2+x**2)**1.5)\n",
+ "a=B1*math.pi*A2**2\n",
+ "M=a/I\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Flux linking the bigger loop is\", round(a*10**11,1)\n",
+ "print\"(ii) Mutual induction between the two loops is\",round(M*10**11,2),\"!0**-11 H\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Flux linking the bigger loop is 9.1\n",
+ "(ii) Mutual induction between the two loops is 4.55 !0**-11 H\n"
+ ]
+ }
+ ],
+ "prompt_number": 181
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 12.26 Page no 688"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "l=0.5 #m\n",
+ "n=20 #turns\n",
+ "r=50 #cm\n",
+ "A1=40*10**-4 #m**2\n",
+ "n1=25\n",
+ "A2=25*10**-4 #m**2\n",
+ "\n",
+ "#Calculation\n",
+ "u=4*math.pi*10**-7\n",
+ "N=n*r\n",
+ "N2=n1*r\n",
+ "M=(u*N*N2*A2)/l\n",
+ "\n",
+ "#Result\n",
+ "print\"Mutual induction of the system is\",round(M*10**3,2),\"*10**-3 H\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Mutual induction of the system is 7.85 *10**-3 H\n"
+ ]
+ }
+ ],
+ "prompt_number": 188
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter13_1.ipynb b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter13_1.ipynb new file mode 100644 index 00000000..8800aa9c --- /dev/null +++ b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter13_1.ipynb @@ -0,0 +1,1642 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:2cfb52e11f318cb94f2cd1051460c5732d92997d5b28187a631460f42051172b"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 13 Alternating currents"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.1 Page no 721"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "I0=141.4 #A\n",
+ "w=314\n",
+ "t=3*10**-3 #s\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "f=w/(2*math.pi)\n",
+ "T=1/f\n",
+ "I=-I0*t*math.sin(314*180/3.14)\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) The maximum value is\",I0,\"A\"\n",
+ "print\"(ii) Frequency is\",round(f,0),\"Hz\"\n",
+ "print\"(iii) Time period is\",round(T,2),\"S\"\n",
+ "print\"(iv) The instantaneous value is\", round(I*10**3,2),\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) The maximum value is 141.4 A\n",
+ "(ii) Frequency is 50.0 Hz\n",
+ "(iii) Time period is 0.02 S\n",
+ "(iv) The instantaneous value is 411.54 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 31
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.3 Page no 721"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "E=220 #V\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "E0=math.sqrt(2)*E\n",
+ "Emean=2*E0/math.pi\n",
+ "\n",
+ "#Result\n",
+ "print\"Average e.m.f during a positive half cycle is\", round(Emean,0),\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Average e.m.f during a positive half cycle is 198.0 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 38
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.4 Page no 721"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "A=2\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "I=math.sqrt(A**2)\n",
+ "\n",
+ "#Result\n",
+ "print\"r.m.s Value of current is\", I,\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "r.m.s Value of current is 2.0 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 42
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.5 Page no 722"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "I0=120 #A\n",
+ "a=360.0\n",
+ "b=96\n",
+ "c=120.0\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "t=1/a\n",
+ "I=I0*math.sin(math.pi/3.0)\n",
+ "a1=b/c\n",
+ "a2=math.asin(a1)\n",
+ "t=a2/(c*math.pi)\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Instantaneous value after 1/360 second is\",round(I,2),\"A\"\n",
+ "print\"(ii) Time taken to reach 96 A for the first time is\", round(t,5),\"S\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Instantaneous value after 1/360 second is 103.92 A\n",
+ "(ii) Time taken to reach 96 A for the first time is 0.00246 S\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.7 Page no 726"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "E0=60\n",
+ "R=20.0 #ohm\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "Ev=E0/(math.sqrt(2))\n",
+ "Iv=Ev/R\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) A.C ammeter will\",round(Iv,2),\"A\"\n",
+ "print\"(ii) Average value of a.c over one cycle is zero\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) A.C ammeter will 2.12 A\n",
+ "(ii) Average value of a.c over one cycle is zero\n"
+ ]
+ }
+ ],
+ "prompt_number": 73
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.8 Page no 726"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "E0=250 #V\n",
+ "I0=10 #A\n",
+ "\n",
+ "#Calculation\n",
+ "P=E0*I0\n",
+ "P1=P/2.0\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Peak power is\", P,\"W\"\n",
+ "print\"(ii) Average power is\",P1,\"W\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Peak power is 2500 W\n",
+ "(ii) Average power is 1250.0 W\n"
+ ]
+ }
+ ],
+ "prompt_number": 78
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.9 Page no 726"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Ev=120.0\n",
+ "P=1000 #W\n",
+ "Ev1=240\n",
+ "\n",
+ "#Calculation\n",
+ "Iv=P/Ev\n",
+ "I0=math.sqrt(2)*Iv\n",
+ "R=Ev/Iv\n",
+ "P=Ev1**2/R\n",
+ "\n",
+ "#Result\n",
+ "print\"Resistance is\", R,\"ohm \\nPeak current is\",P,\"W\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Resistance is 14.4 ohm \n",
+ "Peak current is 4000.0 W\n"
+ ]
+ }
+ ],
+ "prompt_number": 88
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.11 Page no 729"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Xl=220 #ohm\n",
+ "L=0.7 #H\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "f=Xl/(2*math.pi*L)\n",
+ "\n",
+ "#Result\n",
+ "print\"Frequency is\", round(f,0),\"HZ\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Frequency is 50.0 HZ\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.12 Page no 729"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "f=50 #Hz\n",
+ "I=1.4\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "E=2*math.pi*f*I*2*math.cos(2*math.pi*f)\n",
+ "Ev=E/math.sqrt(2)\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Potential difference across the coil is\", round(E,0),\"cos 100*math.pi*t\"\n",
+ "print\"(ii) r.m.s value of p.d across the coil is\", round(Ev,1),\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Potential difference across the coil is 880.0 cos 100*math.pi*t\n",
+ "(ii) r.m.s value of p.d across the coil is 622.0 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 16
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.13 Page no 729"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "f=50 #Hz\n",
+ "L=2 \n",
+ "Ev=12 #V\n",
+ "L1=6\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "Xl=2*math.pi*f*L\n",
+ "Iv=Ev/Xl\n",
+ "Xl1=2*math.pi*f*L1\n",
+ "Iv1=Ev/Xl1\n",
+ "\n",
+ "#Result\n",
+ "print\"Current flows when the inductance is changed to 6 H\", round(Iv1,4),\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Current flows when the inductance is changed to 6 H 0.0064 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 23
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.14 Page no 729"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Ev=200 #V\n",
+ "I0=0.9 #A\n",
+ "f=50 #Hz\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "E0=math.sqrt(2)*Ev\n",
+ "Xl=E0/I0\n",
+ "L=Xl/(2*math.pi*f)\n",
+ "\n",
+ "#Result\n",
+ "print\"The value of inductance is\", round(L,0),\"H\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of inductance is 1.0 H\n"
+ ]
+ }
+ ],
+ "prompt_number": 29
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.15 Page no 732"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "C=318*10**-6 #F\n",
+ "Ev=230 #V\n",
+ "f=50 #Hz\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "Xc=1/(2*math.pi*f*C)\n",
+ "Iv=Ev/Xc\n",
+ "E0=math.sqrt(2)*Ev\n",
+ "I0=math.sqrt(2)*Iv\n",
+ "w=2*math.pi*f\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) The capacitive reactance is\",round(Xc,0),\"ohm\"\n",
+ "print\"(ii) r.m.s value of circuit current is\",round(Iv,0),\"A\"\n",
+ "print\"(iii) Equation for voltage is\",round(E0,2),\"sin 314 t\"\n",
+ "print\"Equation for current is\",round(I0,2),\"sin (314 t+pi/2)\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) The capacitive reactance is 10.0 ohm\n",
+ "(ii) r.m.s value of circuit current is 23.0 A\n",
+ "(iii) Equation for voltage is 325.27 sin 314 t\n",
+ "Equation for current is 32.5 sin (314 t+pi/2)\n"
+ ]
+ }
+ ],
+ "prompt_number": 49
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.16 Page no 732"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "L=1 #H\n",
+ "Xl=3142.0 #ohm\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "f=Xl/(2*math.pi*L)\n",
+ "C=1/(2.0*math.pi*f*Xl)\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Value of frequency is\", round(f,0),\"ohm\"\n",
+ "print\"(ii) Capacity of a condenser is\",round(C*10**6,2),\"micro F\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Value of frequency is 500.0 ohm\n",
+ "(ii) Capacity of a condenser is 0.1 micro F\n"
+ ]
+ }
+ ],
+ "prompt_number": 59
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.17 Page no 732"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "C=50*10**-6 #F\n",
+ "V=230 #V\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "q=C*V*math.sqrt(2)\n",
+ "E=0.5*C*(V*math.sqrt(2))**2\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Maximum charge on the capacitor is\", round(q*10**3,2),\"*10**-3 C\"\n",
+ "print\"(ii) The maximum energy stored in the capacitor is\", round(E,2),\"J\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Maximum charge on the capacitor is 16.26 *10**-3 C\n",
+ "(ii) The maximum energy stored in the capacitor is 2.65 J\n"
+ ]
+ }
+ ],
+ "prompt_number": 69
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.18 Page no 732"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "I0=10 #A\n",
+ "w=314\n",
+ "L=5\n",
+ "\n",
+ "#Calculation\n",
+ "E=0.5*L*I0**2\n",
+ "E0=w*L*I0\n",
+ "C=(E*2)/(E0**2)\n",
+ "\n",
+ "#Result\n",
+ "print\"Capacitance of the capacitor is\",round(C*10**6,2),\"micro F\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Capacitance of the capacitor is 2.03 micro F\n"
+ ]
+ }
+ ],
+ "prompt_number": 76
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.19 Page no 737"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "f=50\n",
+ "L=31.8*10**-3 #H\n",
+ "R=7.0 #ohm\n",
+ "Ev=230 #V\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "Xl=2*math.pi*f*L\n",
+ "Z=math.sqrt(R**2+Xl**2)\n",
+ "Iv=Ev/Z\n",
+ "T=Xl/R\n",
+ "a=math.atan(T)*180/3.14\n",
+ "a1=math.cos(a)*3.14/180.0\n",
+ "P=Iv**2*R\n",
+ "t=55*math.pi/(180.0*3.14)\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Circuit current is\",round(Iv,2),\"A\"\n",
+ "print\"(ii) Phase angle is\", round(a,0),\"lag\"\n",
+ "print\"(iii) Power factor is\", round(a1*10**3,3),\"lag\"\n",
+ "print\"(iv) Power consumed is\",round(P,0),\"W\"\n",
+ "print\"Time lag between voltage maximum and current maximum is\",round(t*10**1,2),\"*10**-3 S\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Circuit current is 18.85 A\n",
+ "(ii) Phase angle is 55.0 lag\n",
+ "(iii) Power factor is 0.554 lag\n",
+ "(iv) Power consumed is 2488.0 W\n",
+ "Time lag between voltage maximum and current maximum is 3.06 *10**-3 S\n"
+ ]
+ }
+ ],
+ "prompt_number": 121
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.20 Page no 737"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "P=400 #W\n",
+ "Ev=250 #V\n",
+ "Iv=2.5 #A\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "a=P/(Ev*Iv)\n",
+ "Z=Ev/Iv\n",
+ "R=Z*a\n",
+ "Xl=math.sqrt(Z**2-R**2)\n",
+ "L=Xl/(2*math.pi*f)\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) The power factor is\",a,\"lag\"\n",
+ "print\"(ii) Resistance of the coil is\", R,\"ohm\"\n",
+ "print\"(iii) Inductance of the coil is\", round(L,3),\"H\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) The power factor is 0.64 lag\n",
+ "(ii) Resistance of the coil is 64.0 ohm\n",
+ "(iii) Inductance of the coil is 0.245 H\n"
+ ]
+ }
+ ],
+ "prompt_number": 134
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.21 Page no 738"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Vr=150 #V\n",
+ "R=75.0 #ohm\n",
+ "f=50 #Hz\n",
+ "L=318*10**-3 #H\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "Iv=Vr/R\n",
+ "Xl=2*math.pi*f*L\n",
+ "Vl=Iv*Xl\n",
+ "Z=math.sqrt(R**2+Xl**2)\n",
+ "Ev=Iv*Z\n",
+ "a=Xl/R\n",
+ "a1=math.atan(a)*180/3.14\n",
+ "\n",
+ "#Result\n",
+ "print \"(i) The supply voltage is\",round(Ev,0),\"V\"\n",
+ "print\"(ii) The phase angle is\",round(a1,2),\"degree lag\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) The supply voltage is 250.0 V\n",
+ "(ii) The phase angle is 53.13 degree lag\n"
+ ]
+ }
+ ],
+ "prompt_number": 149
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.22 Page no 738"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "P=60 #W\n",
+ "Ev=100.0 #V\n",
+ "Ev1=220 #v\n",
+ "f=50 #Hz\n",
+ "\n",
+ "#Calculation\n",
+ "Iv=P/Ev\n",
+ "Vr=Ev1-Ev\n",
+ "R=Vr/Iv\n",
+ "Xl=Vl/Iv\n",
+ "Vl=math.sqrt(Ev1**2-Ev**2)\n",
+ "L=Xl/(2*math.pi*f)\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) The value of non inductive resistance is\", R,\"ohm\"\n",
+ "print\"(ii) Pure inductance is\",round(L,2),\"H\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) The value of non inductive resistance is 200.0 ohm\n",
+ "(ii) Pure inductance is 1.04 H\n"
+ ]
+ }
+ ],
+ "prompt_number": 163
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.23 Page no 739"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "f1=50.0\n",
+ "L=1\n",
+ "E=100 #V\n",
+ "I=1.0 #A\n",
+ "Iv=0.5 #A\n",
+ "f=0\n",
+ "#Calculation\n",
+ "import math\n",
+ "Xl=2*math.pi*f*L\n",
+ "R=E/I\n",
+ "Z=Ev/Iv\n",
+ "Xl1=math.sqrt(Z**2-R**2)\n",
+ "L=Xl1/(2.0*math.pi*f1)\n",
+ "\n",
+ "#Result\n",
+ "print\"The value of resistance is\",R ,\"ohm\"\n",
+ "print\"The value of impedence is\",Z,\"ohm\"\n",
+ "print\"Inductance of the coil is\",round(L,2),\"H\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of resistance is 100.0 ohm\n",
+ "The value of impedence is 200.0 ohm\n",
+ "Inductance of the coil is 0.55 H\n"
+ ]
+ }
+ ],
+ "prompt_number": 183
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.24 Page no 742"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "I=10**6 #A\n",
+ "f=50 #Hz\n",
+ "C=79.5\n",
+ "R=30 #ohm\n",
+ "Ev=100 #V\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "Xc=I/(2*math.pi*f*C)\n",
+ "Z=math.sqrt(R**2+Xc**2)\n",
+ "Iv=Ev/Z\n",
+ "a=Xc/R\n",
+ "a1=math.atan(a)*180/3.14\n",
+ "w=2*math.pi*f\n",
+ "I0=math.sqrt(2)*Iv\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Circuit impedence is\", round(Z,0),\"ohm\"\n",
+ "print\"(ii) Circuit current is\",round(Iv,0),\"A\"\n",
+ "print\"(iii) Phase angle is\",round(a1,0),\"degree lead\"\n",
+ "print\"(iv) Equation for the instantaneous value of current is\",round(I0,3),\"sin(\",round(w,0),\"t+\",round(a1,0),\"degree)\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Circuit impedence is 50.0 ohm\n",
+ "(ii) Circuit current is 2.0 A\n",
+ "(iii) Phase angle is 53.0 degree lead\n",
+ "(iv) Equation for the instantaneous value of current is 2.827 sin( 314.0 t+ 53.0 degree)\n"
+ ]
+ }
+ ],
+ "prompt_number": 24
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.25 Page no 742"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "P=80 #W\n",
+ "V=100.0 #v\n",
+ "V1=200 #V\n",
+ "f=50 #Hz\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "Iv=P/V\n",
+ "Vc=math.sqrt(V1**2-V**2)\n",
+ "Xc=Vc/Iv\n",
+ "C=1/(2.0*math.pi*f*Xc)\n",
+ "\n",
+ "#Result\n",
+ "print\"Capcitance of a capacitor is\", round(C*10**6,1),\"micro F\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Capcitance of a capacitor is 14.7 micro F\n"
+ ]
+ }
+ ],
+ "prompt_number": 32
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.26 Page no 742"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Ev=200 #V\n",
+ "Iv=10.0\n",
+ "f=50 #Hz\n",
+ "\n",
+ "#Calculation\n",
+ "z=Ev/Iv\n",
+ "R=z*math.cos(30*3.14/180.0)\n",
+ "Xc=z*math.sin(30*3.14/180.0)\n",
+ "C=1/(2.0*math.pi*f*Xc)\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Value of resistance is\", round(R,2),\"ohm\"\n",
+ "print\"(ii) Capacitive reactance is\", round(Xc,0),\"ohm\"\n",
+ "print\"(iii) Capacitance of the circuit is\", round(C*10**6,0),\"micro F\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Value of resistance is 17.32 ohm\n",
+ "(ii) Capacitive reactance is 10.0 ohm\n",
+ "(iii) Capacitance of the circuit is 318.0 micro F\n"
+ ]
+ }
+ ],
+ "prompt_number": 47
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.27 Page no 743"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Iv=5 #A\n",
+ "R=10 #ohm\n",
+ "Ev=60 #V\n",
+ "C=400 #micro F\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "Vr=Iv*R\n",
+ "Vc=math.sqrt(Ev**2-Vr**2)\n",
+ "Xc=Vc/Iv\n",
+ "f=1/(2.0*math.pi*C*Xc)\n",
+ "a=Vc/Vr\n",
+ "a1=math.atan(a)*180/3.14\n",
+ "\n",
+ "#Result\n",
+ "print\"The value of supplied frequency is\", round(f*10**6,0),\"Hz\"\n",
+ "print\"Phase angle between circuit current and supply voltage is\",round(a1,1),\"degree lead\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of supplied frequency is 60.0 Hz\n",
+ "Phase angle between circuit current and supply voltage is 33.6 degree lead\n"
+ ]
+ }
+ ],
+ "prompt_number": 65
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.28 Page no 743"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R=200 #ohm\n",
+ "C=15*10**-6 #F\n",
+ "Ev=220 #V\n",
+ "f=50 #Hz\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "Xc=1/(2*math.pi*f*C)\n",
+ "Z=math.sqrt(R**2+Xc**2)\n",
+ "Iv=Ev/Z\n",
+ "Vr=Iv*R\n",
+ "Vc=Iv*Xc\n",
+ "V=Vr+Vc\n",
+ "Vrc=math.sqrt(Vr**2+Vc**2)\n",
+ "\n",
+ "#Result\n",
+ "print\"(a) The current in the circuit is\", round(Iv,3),\"A\"\n",
+ "print\"(b) Voltage across the resistor and capacitor is\",Vrc,\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " (a) The current in the circuit is 0.754 A\n",
+ "(b) Voltage across the resistor and capacitor is 220.0 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 83
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.29 Page no 746"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R1=10.0 #ohm\n",
+ "R2=5.0 #ohm\n",
+ "R3=15 #ohm\n",
+ "Ev=200\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "R=R1+R2+R3\n",
+ "X=R3-(R1+R3)\n",
+ "Z=math.sqrt(R**2+R1**2)\n",
+ "Iv=Ev/Z\n",
+ "T=X/R\n",
+ "a=-math.atan(T)*180/3.14\n",
+ "b=math.cos(a*3.14/180.0)\n",
+ "P=Iv**2*R\n",
+ "print\"(i) Circuit current is\",round(Iv,2) ,\"A\"\n",
+ "print\"(ii) Circuit phase angle is\",round(a,2),\"degree lead\"\n",
+ "print\"(iii)Phase angle between applied voltage and circuit current \",round(b,3),\"lead\"\n",
+ "print\"(iv)Power consumed is\",P,\"W\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Circuit current is 6.32 A\n",
+ "(ii) Circuit phase angle is 18.44 degree lead\n",
+ "(iii)Phase angle between applied voltage and circuit current 0.949 lead\n",
+ "(iv)Power consumed is 1200.0 W\n"
+ ]
+ }
+ ],
+ "prompt_number": 116
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.30 Page no. 746"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "F=50 #HZ\n",
+ "L=0.06 \n",
+ "C=6.8\n",
+ "l=10**6\n",
+ "R=2.5\n",
+ "Ev=230 #V\n",
+ "\n",
+ "#Calculation\n",
+ "import math \n",
+ "Xl=2*math.pi*F*L\n",
+ "Xc=l/(2*math.pi*F*C)\n",
+ "Z=math.sqrt(R**2+(Xl-Xc)**2)\n",
+ "Iv=Ev/Z\n",
+ "a=(Xl-Xc)/R\n",
+ "a2=-math.atan(a)*180.0/3.14\n",
+ "P=R/Z\n",
+ "P1=Ev*Iv*P\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Circuit impedence is\",round(Z,1),\"ohm\"\n",
+ "print\"(ii) Circuit current is\",round(Iv,3),\"A\"\n",
+ "print\"(iii) Phase angle is \",round(a2,1),\"degree lead\" \n",
+ "print\"(iv) Power factor is\",round(P,4),\"lead\"\n",
+ "print\"(v) Power consumed is\",round(P1,2),\"W\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Circuit impedence is 449.3 ohm\n",
+ "(ii) Circuit current is 0.512 A\n",
+ "(iii) Phase angle is 89.7 degree lead\n",
+ "(iv) Power factor is 0.0056 lead\n",
+ "(v) Power consumed is 0.66 W\n"
+ ]
+ }
+ ],
+ "prompt_number": 146
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.31 Page no. 746"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=65 #degree\n",
+ "b=20 #degree\n",
+ "w=3000\n",
+ "L=0.01\n",
+ "E0=400 #V\n",
+ "I=10\n",
+ "f=50\n",
+ "\n",
+ "#calculation\n",
+ "import math\n",
+ "a=a-b\n",
+ "Xl=w*L\n",
+ "Z=E0/(I*math.sqrt(2))\n",
+ "R=Z/math.sqrt(2)\n",
+ "Xc=Xl-R\n",
+ "C=1/(w*Xc*10**-6)\n",
+ "\n",
+ "#Result\n",
+ "print\" The value of C is\" ,round(C,1),\"microF\"\n",
+ "print\" The value of R is\",R,\"ohm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " The value of C is 33.3 microF\n",
+ " The value of R is 20.0 ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 161
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.32 Page no 747"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "f=50 #Hz\n",
+ "L=0.03\n",
+ "R=8 #ohm\n",
+ "Ev=240 #V\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "Xl=2*math.pi*f*L\n",
+ "Z=math.sqrt(R**2+Xl**2)\n",
+ "Iv=Ev/Z\n",
+ "P=Iv**2*R\n",
+ "a=R/Z\n",
+ "Xc=2*Xl\n",
+ "C=1/(2*math.pi*f*Xc)\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) The value of current is\",round(Iv,2) ,\"A\"\n",
+ "print\" The value of power is\",round(P,0),\"W\"\n",
+ "print \" Power factor is\",round(a,2),\"lag\"\n",
+ "print\"(ii) The vaue of capacitance is\", round(C*10**6,0),\"micro F\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) The value of current is 19.41 A\n",
+ " The value of power is 3015.0 W\n",
+ " Power factor is 0.65 lag\n",
+ "(ii) The vaue of capacitance is 169.0 micro F\n"
+ ]
+ }
+ ],
+ "prompt_number": 186
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.33 Page no 747"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Vr=65 #V\n",
+ "R=100.0 #ohm\n",
+ "Vl=204\n",
+ "Vc=415\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "Iv=Vr/R\n",
+ "Xl=Vl/Iv\n",
+ "L=Xl/(2*math.pi*f)\n",
+ "Xc=Vc/Iv\n",
+ "C=1/(2*math.pi*f*Xc)\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Circuit current is\", Iv,\"A\"\n",
+ "print\"(ii) Inductance is\",round(L,0),\"H\"\n",
+ "print\"(iii) The value of capacitance is\",round(C*10**6,0),\"micro F\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Circuit current is 0.65 A\n",
+ "(ii) Inductance is 1.0 H\n",
+ "(iii) The value of capacitance is 5.0 micro F\n"
+ ]
+ }
+ ],
+ "prompt_number": 201
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.34 Page no 752"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "C=100*10**-12 #F\n",
+ "L=100*10**-6 #H\n",
+ "Ev=10\n",
+ "R=100.0 #ohm\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "fr=1/(2*math.pi*math.sqrt(L*C))\n",
+ "Iv=Ev/R\n",
+ "Vl=Iv*2*math.pi*fr*L\n",
+ "Vc=Iv/(2.0*math.pi*fr*C)\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Resonant frequency is\", round(fr*10**-6,2),\"*10**6 HZ\"\n",
+ "print\"(ii) Current at resonance is\", Iv,\"A\"\n",
+ "print\"(iii) Voltage across L and C is\", Vc,\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Resonant frequency is 1.59 *10**6 HZ\n",
+ "(ii) Current at resonance is 0.1 A\n",
+ "(iii) Voltage across L and C is 100.0 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 218
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.35 Page no 752"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "f=50 #Hz\n",
+ "L=0.5\n",
+ "Ev=100 #v\n",
+ "R=4 #ohm\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "C=1/(4*math.pi**2*f**2*L)\n",
+ "Ir=Ev/R\n",
+ "Vr=Ir*2*math.pi*f*L\n",
+ "Vc=Ir/(2*math.pi*f*C)\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) The capacitance is\", round(C*10**6,2),\"micro F\"\n",
+ "print\"(ii) The voltage across inductance and capacitance is\", round(Vc,0),\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) The capacitance is 20.26 micro F\n",
+ "(ii) The voltage across inductance and capacitance is 3927.0 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 229
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.36 Page no 752"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "f=50 #Hz\n",
+ "L=0.318 #H\n",
+ "Iv=2.3\n",
+ "R=100 #ohm\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "C=1/((2*math.pi*f)**2*L)\n",
+ "Vl=Iv*2*math.pi*f*C*10**4\n",
+ "P=Iv**2*R\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) The value of capacitor is\", round(C*10**6,1),\"micro F\"\n",
+ "print\"(ii) Voltage across the inductor is\", round(Vl,0),\"V\"\n",
+ "print\"(iii)Total power consumed is\",P,\"W\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) The value of capacitor is 31.9 micro F\n",
+ "(ii) Voltage across the inductor is 230.0 V\n",
+ "(iii)Total power consumed is 529.0 W\n"
+ ]
+ }
+ ],
+ "prompt_number": 245
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.37 Page no 753"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "E0=283 #V\n",
+ "f=50 #Hz\n",
+ "R=3.0 #ohm\n",
+ "L=25.48*10**-3 #h\n",
+ "C=796*10**-6 #F\n",
+ "Xl=8\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "Xc=1/(2*math.pi*f*C)\n",
+ "Z=math.sqrt(R**2+(Xl-Xc)**2)\n",
+ "a=math.atan(Xc/R)*180/3.14\n",
+ "Iv=(E0/math.sqrt(2))/Z\n",
+ "P=Iv**2*R\n",
+ "a1=math.cos(a*180/3.14)\n",
+ "\n",
+ "#Result\n",
+ "print\"(a) The inpedence of the circuit is\", round(Z,0),\"ohm\"\n",
+ "print\"(b) The phase difference is\", round(a,1),\"degree\"\n",
+ "print\"(c) The power dissipated is\", round(P,0),\"W\"\n",
+ "print\"(d) Power factor is\", round(a1,1),\"lag\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) The inpedence of the circuit is 5.0 ohm\n",
+ "(b) The phase difference is 53.1 degree\n",
+ "(c) The power dissipated is 4804.0 W\n",
+ "(d) Power factor is 0.8 lag\n"
+ ]
+ }
+ ],
+ "prompt_number": 270
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.38 Page no 753"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "L=25.48*10**-3 #H\n",
+ "C=796*10**-6\n",
+ "R=3.0 #ohm\n",
+ "E0=283\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "fr=1/(2.0*math.pi*math.sqrt(L*C))\n",
+ "Iv=(E0/math.sqrt(2))/R\n",
+ "P=Iv**2*R\n",
+ "\n",
+ "#Result\n",
+ "print\"(a) Frequency of the source is\", round(fr,1),\"Hz\"\n",
+ "print\"(b) The value of impedence is\",R,\"ohm\"\n",
+ "print\"The value of current is\",round(Iv,1),\"A\"\n",
+ "print\"The power dissipated is\",round(P,0),\"W\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) Frequency of the source is 35.3 Hz\n",
+ "(b) The value of impedence is 3.0 ohm\n",
+ "The value of current is 66.7 A\n",
+ "The power dissipated is 13348.0 W\n"
+ ]
+ }
+ ],
+ "prompt_number": 287
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.39 Page no 757"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "C=1200*10**-12 #F\n",
+ "E=500\n",
+ "L=0.075 #H\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "q0=C*E\n",
+ "I0=q0/(math.sqrt(L*C))\n",
+ "f=1/(2*math.pi*math.sqrt(L*C))\n",
+ "T=1/f\n",
+ "U=q0**2/(2.0*C)\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) The initial charge onthe capcitor is\",q0,\"c\"\n",
+ "print\"(ii) The maximum current is\",round(I0*10**3,0),\"mA\"\n",
+ "print\"(iii) The value of frequency is\", round(f*10**-3,0),\"*10**3 Hz\"\n",
+ "print\"Time period is\", round(T*10**5,0),\"*10**-5 S\"\n",
+ "print\"(iv) Total energy is\",U*10**4,\"*10**-4 J\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) The initial charge onthe capcitor is 6e-07 c\n",
+ "(ii) The maximum current is 63.0 mA\n",
+ "(iii) The value of frequency is 17.0 *10**3 Hz\n",
+ "Time period is 6.0 *10**-5 S\n",
+ "(iv) Total energy is 1.5 *10**-4 J\n"
+ ]
+ }
+ ],
+ "prompt_number": 315
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.40 Page no 758"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "L=8*10**-6 #H\n",
+ "C=0.02*10**-6 #F\n",
+ "c=3*10**8\n",
+ "\n",
+ "#Calculation\n",
+ "f=1/(2*math.pi*math.sqrt(L*C))\n",
+ "w=c/f\n",
+ "\n",
+ "#Result\n",
+ "print\"Wavelength is\", round(w*10**-2,2),\"*10**2 m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Wavelength is 7.54 *10**2 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 321
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter14_1.ipynb b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter14_1.ipynb new file mode 100644 index 00000000..7bb51839 --- /dev/null +++ b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter14_1.ipynb @@ -0,0 +1,629 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:1bd22700738f4a2a80468f70e18e63c26ad56b1a125cfb69784af1d3eb280a8a"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 14 Electrical devices"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 14.1 Page no 787"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "N=100\n",
+ "A=10**-2 #m**2\n",
+ "B=0.5 #T\n",
+ "f=500/60.0\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "w=2*math.pi*f\n",
+ "E0=N*A*B*w\n",
+ "E=E0*math.sin(60*3.14/180.0)\n",
+ "\n",
+ "#Result\n",
+ "print\"Maximum emf produced in the coil is\", round(E0,2),\"V\"\n",
+ "print\"Instantaneous value of e.m.f. is\",round(E,1),\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Maximum emf produced in the coil is 26.18 V\n",
+ "Instantaneous value of e.m.f. is 22.7 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 14.2 Page no 787"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "N=50\n",
+ "A=2.5\n",
+ "B=0.3 #T\n",
+ "w=60\n",
+ "R=500 #ohm\n",
+ "\n",
+ "#Calculation\n",
+ "E0=N*B*A*w\n",
+ "I0=E0/R\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Maximum current drawn from the gnerator is\",I0,\"A\"\n",
+ "print\"(ii) The maximum flux through the coil is zero and when induced current is maximum, flux through the coil is zero\"\n",
+ "print\"(iii) Yes the generator will work because the basic condition for induced an .m.f. is that there should be relative motion between coil and magnetic field\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Maximum current drawn from the gnerator is 4.5 A\n",
+ "(ii) The maximum flux through the coil is zero and when induced current is maximum, flux through the coil is zero\n",
+ "(iii) Yes the generator will work because the basic condition for induced an .m.f. is that there should be relative motion between coil and magnetic field\n"
+ ]
+ }
+ ],
+ "prompt_number": 17
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 14.3 Page no 787"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "N=150\n",
+ "A=2*10**-2 #m**2\n",
+ "B=0.15 #T\n",
+ "f=60\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "w=2*math.pi*f\n",
+ "E0=N*A*B*w\n",
+ "\n",
+ "#Result\n",
+ "print\"Peak value of e.m.f is\", round(E0,0),\"V\"\n",
+ "print\"Average value of induced e.m.f is zero\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Peak value of e.m.f is 170.0 V\n",
+ "Average value of induced e.m.f is zero\n"
+ ]
+ }
+ ],
+ "prompt_number": 23
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 14.4 Page no 787"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "N=100\n",
+ "A=3\n",
+ "B=0.04 #T\n",
+ "w=60\n",
+ "R=500 #ohm\n",
+ "\n",
+ "#Calculation\n",
+ "E0=N*A*B*w\n",
+ "I0=E0/R\n",
+ "P=E0*I0\n",
+ "\n",
+ "#Result\n",
+ "print\"Maximum power dissipated in the coil is\", P,\"W\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Maximum power dissipated in the coil is 1036.8 W\n"
+ ]
+ }
+ ],
+ "prompt_number": 28
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 14.5 Page no 788"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "N=100\n",
+ "A=0.10 #m**2\n",
+ "f=0.5 #Hz\n",
+ "B=0.01 #T\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "w=2*math.pi*f\n",
+ "E0=N*A*B*w\n",
+ "\n",
+ "#Result\n",
+ "print\"Maximum voltage generated in the coil is\", round(E0,3),\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Maximum voltage generated in the coil is 0.314 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 32
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 14.6 Page no 792"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "V=240 #V\n",
+ "I=5 #A\n",
+ "R=4 #ohm\n",
+ "\n",
+ "#Calculation\n",
+ "Eb=V-(I*R)\n",
+ "\n",
+ "#Result\n",
+ "print\"Value of back e.m.f is\", Eb,\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Value of back e.m.f is 220 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 35
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 14.7 Page no 792"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "I=20 #A\n",
+ "R=2 #ohm\n",
+ "n=0.5 \n",
+ "P=2000 #W\n",
+ "\n",
+ "#Calculation\n",
+ "P1=P/n\n",
+ "V=P1/I\n",
+ "Eb=V-(I*R)\n",
+ "\n",
+ "#Result\n",
+ "print\"The back e.m.f is\", Eb,\"V \\nSupply voltage is\",V,\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The back e.m.f is 160.0 V \n",
+ "Supply voltage is 200.0 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 44
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 14.8 Page no 793"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "V=100 #V\n",
+ "I=6 #A\n",
+ "V1=0.7\n",
+ "\n",
+ "#Calculation\n",
+ "Pin=V*I\n",
+ "R=(V1*Pin)/I**2\n",
+ "\n",
+ "#Result\n",
+ "print\"Armature resistance is\", round(R,2),\"ohm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Armature resistance is 11.67 ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 49
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 14.10 Page no 793"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "V=200 #V\n",
+ "I=5 #A\n",
+ "R=8.5 #ohm \n",
+ "\n",
+ "#Calculation\n",
+ "Eb=V-(I*R)\n",
+ "Pi=V*I\n",
+ "P0=Eb*I\n",
+ "n=(P0*100)/Pi\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Back e.m.f of motor is\", Eb,\"V\"\n",
+ "print\"(ii) Power input is\",Pi,\"W\"\n",
+ "print\"(iii) Output power is\",P0,\"W\"\n",
+ "print\"(iv) Efficiency of motor is\",n,\"%\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Back e.m.f of motor is 157.5 V\n",
+ "(ii) Power input is 1000 W\n",
+ "(iii) Output power is 787.5 W\n",
+ "(iv) Efficiency of motor is 78.75 %\n"
+ ]
+ }
+ ],
+ "prompt_number": 61
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 14.11 Page no 796"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Vp=200 #V\n",
+ "n=200.0\n",
+ "Ip=2 #A\n",
+ "\n",
+ "#Calculation\n",
+ "Vs=Vp*n\n",
+ "Is=(Ip*V)/Vs\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Voltage developed in the secondary is\", Vs,\"V\"\n",
+ "print\"(ii) The current in the secondary is\",Is ,\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Voltage developed in the secondary is 40000.0 V\n",
+ "(ii) The current in the secondary is 0.01 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 69
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 14.12 Page no 796"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Vp=220.0 #V\n",
+ "Is=5 #A\n",
+ "n=20\n",
+ "\n",
+ "#Calculation\n",
+ "Vs=Vp*n\n",
+ "Ip=(Vs*Is)/Vp\n",
+ "P=Vs*Is\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Voltage across secondary is\",Vs,\"V\"\n",
+ "print\"(ii) The current in primary is\",Ip,\"A\"\n",
+ "print\"(iii) The power output is\",P*10**-3,\"K W\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Voltage across secondary is 4400.0 V\n",
+ "(ii) The current in primary is 100.0 A\n",
+ "(iii) The power output is 22.0 K W\n"
+ ]
+ }
+ ],
+ "prompt_number": 79
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 14.13 Page no 797"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "P=120*10**3 #W\n",
+ "R=0.4 #ohm\n",
+ "Ev=240.0 #V\n",
+ "Ev1=24000.0 #V\n",
+ "\n",
+ "#Calculation\n",
+ "Iv=P/Ev\n",
+ "P=Iv**2*R\n",
+ "Iv1=P/Ev1\n",
+ "P1=Iv1**2*R\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Power loss at 240 V is\", P*10**-3,\"K W\"\n",
+ "print\"(ii) Power loss at 24000 V is\", round(P1,0),\"W\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Power loss at 240 V is 100.0 K W\n",
+ "(ii) Power loss at 24000 V is 7.0 W\n"
+ ]
+ }
+ ],
+ "prompt_number": 94
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 14.14 Page no 797"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Np=5000\n",
+ "Vp=2200 #V\n",
+ "Vs=220 #V\n",
+ "Pout=8 #K W\n",
+ "n=0.9\n",
+ "\n",
+ "#Calculation\n",
+ "Ns=(Vs*Np)/Vp\n",
+ "Pin=Pout/n\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) The number of turns in the secondary is\", Ns\n",
+ "print\"(ii) Input power is\",round(Pin,1),\"K W\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) The number of turns in the secondary is 500\n",
+ "(ii) Input power is 8.9 K W\n"
+ ]
+ }
+ ],
+ "prompt_number": 99
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 14.15 Page no 797"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Vp=220.0 #V\n",
+ "Vs=22 #V\n",
+ "Z=220 #ohm\n",
+ "Is=0.1\n",
+ "\n",
+ "#Calclation\n",
+ "Ip=(Vs*Is)/Vp\n",
+ "\n",
+ "#Result\n",
+ "print\"Current drawn is\", Ip,\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Current drawn is 0.01 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 104
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 14.16 Page no 798"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Vs=24 #v\n",
+ "R=9.6 #ohm\n",
+ "Vp=120.0\n",
+ "\n",
+ "#Calculation\n",
+ "Is=Vs/R\n",
+ "Ip=(Vs*Is)/Vp\n",
+ "P1=Vs*Is\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Current in the secondary coil is\", Is,\"A\"\n",
+ "print\"(ii) Current in primary coil is\",Ip ,\"A\"\n",
+ "print\"Power used is\",P1,\"W\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Current in the secondary coil is 2.5 A\n",
+ "(ii) Current in primary coil is 0.5 A\n",
+ "Power used is 60.0 W\n"
+ ]
+ }
+ ],
+ "prompt_number": 116
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter15_1.ipynb b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter15_1.ipynb new file mode 100644 index 00000000..111cd390 --- /dev/null +++ b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter15_1.ipynb @@ -0,0 +1,381 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:977816390da876a89acf9dab4a43ac1149d2a8f7e4f57b74a89090971103e376"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 15 Electromagnetic waves"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 15.1 Page no 836"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "e=8.854*10**-12 #C**2/N/m**2\n",
+ "A=10**-4 #m**2\n",
+ "E=3*10**6 #V/ms\n",
+ "\n",
+ "#Calculation\n",
+ "Id=e*A*E\n",
+ "\n",
+ "#Result\n",
+ "print\"Displacement current is\", round(Id*10**9,1)*10**-9,\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Displacement current is 2.7e-09 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 15.2 Page no 836"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Id=1 #A\n",
+ "C=10.0**-6 #F\n",
+ "\n",
+ "#Calculation\n",
+ "V=Id/C\n",
+ "\n",
+ "#Result\n",
+ "print\"Instantaneous current is\", V,\"V/S\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Instantaneous current is 1000000.0 V/S\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 15.3 Page no 836"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "I=0.15 #A\n",
+ "R=0.12 #m\n",
+ "r=0.065 #m\n",
+ "r1=0.15 #m\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "A=math.pi*R**2\n",
+ "u=4*math.pi*10**-7\n",
+ "B=(u*I*r)/(2*math.pi*R**2)\n",
+ "B1=(u*I)/(2*math.pi*r1)\n",
+ "Bmax=(u*I)/(2*math.pi*R)\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) (a) Magnetic field on the axis is zero\"\n",
+ "print\"(b) Magnetic field at r=6.5 cm is\",round(B*10**7,2)*10**-7 ,\"T\"\n",
+ "print\"(c) Magnetic field at r=15 cm is\", B1,\"T\"\n",
+ "print\"(ii) Distance is\", Bmax,\"T\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) (a) Magnetic field on the axis is zero\n",
+ "(b) Magnetic field at r=6.5 cm is 1.35e-07 T\n",
+ "(c) Magnetic field at r=15 cm is 2e-07 T\n",
+ "(ii) Distance is 2.5e-07 T\n"
+ ]
+ }
+ ],
+ "prompt_number": 24
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 15.5 Page no 837"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "r=0.05 #m\n",
+ "E=10**12 #V/m/s\n",
+ "e=8.854*10**-12\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "Id=e*math.pi*r**2*E\n",
+ "\n",
+ "#Result\n",
+ "print\"Displacement current is\", round(Id,4),\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Displacement current is 0.0695 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 29
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 15.7 Page no 846 "
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "E=100 #v\n",
+ "c=3.0*10**8\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "B=E/c\n",
+ "u=4.0*math.pi*10**-7\n",
+ "H=B/u\n",
+ "U=e*E**2\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Value of B is\", round(B*10**7,2)*10**-7,\" T\"\n",
+ "print\"(ii) Value of H is\",round(H,3),\"A/m\"\n",
+ "print\"(iii) Energy density is\",U,\"J/m**3"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Value of B is 3.33e-07 T\n",
+ "(ii) Value of H is 0.265 A/m\n",
+ "(iii) Energy density is 8.854e-08\n"
+ ]
+ }
+ ],
+ "prompt_number": 41
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 15.8 Page no 847"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "E0=8*10**-4 #v\n",
+ "c=3.0*10**8\n",
+ "w=6*10**6\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "B0=E0/c\n",
+ "f=w/(2.0*math.pi)\n",
+ "l=c/f\n",
+ "\n",
+ "#Result\n",
+ "print\"Wavelength of the wave is\", round(l*10**-4,4),\"m\"\n",
+ "print\"Frequency is\",round(f*10**-6,3),\"*10**10 Hz\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Wavelength of the wave is 0.0314 m\n",
+ "Frequency is 0.955 *10**10 Hz\n"
+ ]
+ }
+ ],
+ "prompt_number": 56
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 15.9 Page no 847"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "E=6.3 #V/m\n",
+ "c=3.0*10**8\n",
+ "\n",
+ "#Calculation\n",
+ "B=E/c\n",
+ "\n",
+ "#Result\n",
+ "print\"B=\", B,\"K^ Tesla\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "B= 2.1e-08 K^ Tesla\n"
+ ]
+ }
+ ],
+ "prompt_number": 59
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 15.10 Page no 847"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "f=18 #W/cm**2\n",
+ "A=20 #cm**2\n",
+ "t=30*60\n",
+ "\n",
+ "#Calculation\n",
+ "U=f*A*t\n",
+ "P=U/c\n",
+ "F=P/t\n",
+ "P1=2*P\n",
+ "F1=P1/t\n",
+ "\n",
+ "#Result\n",
+ "print\"Average force exerted on the surface is\", F1,\"N\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Average force exerted on the surface is 2.4e-06 N\n"
+ ]
+ }
+ ],
+ "prompt_number": 65
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 15.11 Page no 847"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "r=3.0 #m\n",
+ "n=0.025\n",
+ "P=100 #w\n",
+ "C=3*10**8\n",
+ "\n",
+ "#Calculation \n",
+ "import math\n",
+ "A=4*math.pi*r**2\n",
+ "I=(n*P)/A\n",
+ "E0=math.sqrt((2*I)/(e*C))\n",
+ "B0=E0/C\n",
+ "\n",
+ "#Result\n",
+ "print\"Peak value of electric field is\", round(E0,2),\"V/m\"\n",
+ "print\"Peak value of magnetic field is\",round(B0*10**8,2)*10**-8,\"T\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Peak value of electric field is 4.08 V/m\n",
+ "Peak value of magnetic field is 1.36e-08 T\n"
+ ]
+ }
+ ],
+ "prompt_number": 76
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter16_1.ipynb b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter16_1.ipynb new file mode 100644 index 00000000..f74b4c1b --- /dev/null +++ b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter16_1.ipynb @@ -0,0 +1,369 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:badd30ad9356c0b0c69f3c0e035f97e954d74c789af08b85cb880fbe7e95ff6b"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 16 Reflection of light"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Exaple 16.1 Page no 890"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "HF=2\n",
+ "EF=1.9\n",
+ "\n",
+ "#Calculation\n",
+ "L=0.5*HF\n",
+ "DB=0.5*EF\n",
+ "\n",
+ "#Result\n",
+ "print\"Maximum height of mirror is\", L,\"m\"\n",
+ "print\"Bottom edge of the mirror is\",DB,\"m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Maximum height of mirror is 1.0 m\n",
+ "Bottom edge of the mirror is 0.95 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 16.2 Page no 890"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "u=-15.0 #cm\n",
+ "f=-10 #cm\n",
+ "o=2.0 #cm\n",
+ "\n",
+ "#Calculation\n",
+ "v=1/((1.0/f)-(1.0/u))\n",
+ "m=v/u\n",
+ "I=-m*o\n",
+ "\n",
+ "#Result\n",
+ "print\"Position of the image is\", v,\"cm\"\n",
+ "print\"Size of the image is\",I,\"cm\"\n",
+ "print\"Nature of the image isreal and inverted \""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Position of the image is -30.0 cm\n",
+ "Size of the image is -4.0 cm\n",
+ "Nature of the image isreal and inverted \n"
+ ]
+ }
+ ],
+ "prompt_number": 14
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 16.3 Page no 891"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "u=-10.0 #cm\n",
+ "f=-15.0\n",
+ "\n",
+ "#Calculation\n",
+ "v=1/((1/f)-(1/u))\n",
+ "m=-v/u\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Image position is\", v,\"cm\"\n",
+ "print\"(ii) Magnification is\", m"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Image position is 30.0 cm\n",
+ "(ii) Magnification is 3.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 25
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 16.4 Page no 891"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "f=12.0\n",
+ "v=4.0\n",
+ "\n",
+ "#Calculation\n",
+ "u=1/((1/f)-(1/v))\n",
+ "m=-v/u\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Object position is\", u,\"cm\"\n",
+ "print\"(ii) Magnification is\", round(m,2)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Object position is -6.0 cm\n",
+ "(ii) Magnification is 0.67\n"
+ ]
+ }
+ ],
+ "prompt_number": 31
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 16.5 Page no 891"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R=36 #ohm\n",
+ "\n",
+ "#Calculation\n",
+ "f=-R/2.0\n",
+ "u=(2*f)/3.0\n",
+ "\n",
+ "#Result\n",
+ "print\"Position of the object is\", u,\"cm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Position of the object is -12.0 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 36
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 16.6 Page no 892"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R=20 #cm\n",
+ "\n",
+ "#Calculation\n",
+ "f=R/2.0\n",
+ "u=-f\n",
+ "v=-u/2.0\n",
+ "\n",
+ "#Result\n",
+ "print\"Position of the object is\",u,\"cm\"\n",
+ "print\"Position of the image is\",v,\"cm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Position of the object is -10.0 cm\n",
+ "Position of the image is 5.0 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 40
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 16.7 Page no 892"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "f=-15.0\n",
+ "\n",
+ "#Calculation\n",
+ "u=(1/(1/f)/3.0)*4\n",
+ "v=u/2.0\n",
+ "\n",
+ "#Result\n",
+ "print\"Position of object is\" ,u,\"cm\"\n",
+ "print\"When the image is virtual\",v,\"cm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Position of object is -20.0 cm\n",
+ "When the image is virtual -10.0 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 49
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 16.8 Page no 892"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R=30 #ohm\n",
+ "u=-10.0 \n",
+ "h1=5\n",
+ "\n",
+ "#Calculation\n",
+ "f=R/2.0\n",
+ "v=1/((1/f)-(1/u))\n",
+ "h2=(-v*h1)/u\n",
+ "\n",
+ "#Result\n",
+ "print\"Position of the image is\", v,\"cm\"\n",
+ "print\"Size of the image is\",h2,\"cm\"\n",
+ "print\"The image is virtual\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Position of the image is 6.0 cm\n",
+ "Size of the image is 3.0 cm\n",
+ "The image is virtual\n"
+ ]
+ }
+ ],
+ "prompt_number": 58
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 16.9 Page no 893"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "f=-10.0 #cm\n",
+ "u=-25.0 #cm\n",
+ "h1=3\n",
+ "\n",
+ "#Calculation\n",
+ "v=1/((1/f)-(1/u))\n",
+ "h2=(-v*h1)/u\n",
+ "A=h2**2\n",
+ "\n",
+ "#Result\n",
+ "print\"Area enclosed by the image of the wire is\", A,\"cm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Area enclosed by the image of the wire is 4.0 cm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 66
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter17_1.ipynb b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter17_1.ipynb new file mode 100644 index 00000000..6ddbfa96 --- /dev/null +++ b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter17_1.ipynb @@ -0,0 +1,1299 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:7117da667e9242c9a19d8eb5f355fd755219357cacc372784f1e3ef0c539e46b"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 17 Refraction of the light"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 17.1 Page no 918"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "u1=1.50\n",
+ "u2=1.33\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "sinr=u1*math.sin(50*3.14/180.0)/u2\n",
+ "a=math.asin(sinr)*180/3.14\n",
+ "\n",
+ "#Result\n",
+ "print\"Angle of refraction is\", round(a,1),\"degree\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Angle of refraction is 59.8 degree\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 17.2 Page no 918"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "u1=1.0\n",
+ "u2=1.526\n",
+ "i=45 #degree\n",
+ "#Calculation\n",
+ "sinr=(u1*math.sin(i*3.14/180.0))/u2\n",
+ "r=math.asin(sinr)*180/3.14\n",
+ "d=i-r\n",
+ "\n",
+ "#Result\n",
+ "print\"Angle of deviation is\", round(d,2),\"degree\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Angle of deviation is 17.39 degree\n"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 17.3 Page no 918"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "c=3.0*10**8\n",
+ "u=1.5\n",
+ "f=6*10**14 #Hz\n",
+ "\n",
+ "#Calculation\n",
+ "v=c/u\n",
+ "l=c/f\n",
+ "lm=v/f\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Wavelength of light in air is\", l,\"m\"\n",
+ "print\"(ii) Wavelength of light in glass is\",round(lm*10**7,1)*10**-7,\"m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Wavelength of light in air is 5e-07 m\n",
+ "(ii) Wavelength of light in glass is 3.3e-07 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 22
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 17.4 Page no 919"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "ug=1.5\n",
+ "uw=1.3\n",
+ "vw=2.25*10**8\n",
+ "\n",
+ "#Calculation\n",
+ "vg=(uw*vw)/ug\n",
+ "\n",
+ "#Result\n",
+ "print\"Speed of the light in glass is\", vg*10**-8,\"*10**8 m/s\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Speed of the light in glass is 1.95 *10**8 m/s\n"
+ ]
+ }
+ ],
+ "prompt_number": 26
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 17.5 Page no 919"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "u=1.6\n",
+ "t=8\n",
+ "t1=4.5\n",
+ "u1=1.5\n",
+ "t2=6\n",
+ "u2=1.33\n",
+ "\n",
+ "#Calculation\n",
+ "d=t*(1-(1/u))\n",
+ "d1=t1*(1-(1/u1))\n",
+ "d2=t2*(1-(1/u2))\n",
+ "D=d+d1+d2\n",
+ "\n",
+ "#Result\n",
+ "print\"Position of mark from the bottom is\", round(D,0),\"cm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Position of mark from the bottom is 6.0 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 39
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 17.6 Page no 919"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "uw=1.33\n",
+ "uo=1.20\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "uow=uw/uo\n",
+ "sinr=(math.sin(30*3.14/180.0))/uow\n",
+ "r=math.asin(sinr)*180/3.14\n",
+ "\n",
+ "#Result\n",
+ "print\"Angle of refraction in water is\", round(r,1),\"degree\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Angle of refraction in water is 26.8 degree\n"
+ ]
+ }
+ ],
+ "prompt_number": 49
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 17.7 Page no 920"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "v=2.0*10**8 #m/s\n",
+ "c=3*10**8 #m/s\n",
+ "d=6.0 #cm\n",
+ "\n",
+ "#Calculation\n",
+ "ug=c/v\n",
+ "a=d/ug\n",
+ "D=d-a\n",
+ "\n",
+ "#Result\n",
+ "print\"Distance through which ink dot appears to be raised is\", D,\"cm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Distance through which ink dot appears to be raised is 2.0 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 54
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 17.8 Page no 924"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "ug=1.5\n",
+ "uw=1.33\n",
+ "\n",
+ "#Calculation\n",
+ "u1=ug/uw\n",
+ "sinC=1/u1\n",
+ "C=math.asin(sinC)*180/3.14\n",
+ "\n",
+ "#Result\n",
+ "print\"Critical angle is\", round(C,2),\"degree\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Critical angle is 62.49 degree\n"
+ ]
+ }
+ ],
+ "prompt_number": 60
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 17.9 Page no 924"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "v=1.5*10**8\n",
+ "c=3.0*10**8\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "a=v/c\n",
+ "C=math.asin(a)*180/3.14\n",
+ "\n",
+ "#Result\n",
+ "print\"Value of critical angle is\", round(C,0),\"Degree\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Value of critical angle is 30.0 Degree\n"
+ ]
+ }
+ ],
+ "prompt_number": 65
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 17.10 Page no 924"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "uw=1.33\n",
+ "\n",
+ "#Calculation\n",
+ "a=1/uw\n",
+ "b=math.sin(a)*180/3.14\n",
+ "\n",
+ "#Result\n",
+ "print\"Angle of refraction is\", round(b,0),\"degree\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Angle of refraction is 39.0 degree\n"
+ ]
+ }
+ ],
+ "prompt_number": 74
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 17.11 Page no 924"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=4\n",
+ "b=6.0\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "A=a/b\n",
+ "B=math.atan(A)*180/3.14\n",
+ "ur=1/(math.sin(B*3.14/180.0))\n",
+ "\n",
+ "#Result\n",
+ "print\"Refrective index of the liquid is\", round(ur,1)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Refrective index of the liquid is 1.8\n"
+ ]
+ }
+ ],
+ "prompt_number": 79
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 17.12 Page no 925"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=52 #Degree\n",
+ "b=33 #Degree\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "u2=(math.sin(a*3.14/180.0))/(math.sin(b*3.14/180.0))\n",
+ "C=1/u2\n",
+ "A=math.asin(C)*180/3.14\n",
+ "\n",
+ "#Result\n",
+ "print\"Angle of refrection is\", round(A,1),\"Degree\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Angle of refrection is 43.7 Degree\n"
+ ]
+ }
+ ],
+ "prompt_number": 88
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 17.13 Page no 932"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "u=-240.0\n",
+ "R=15.0 #cm\n",
+ "u1=1.33\n",
+ "u2=1.5\n",
+ "\n",
+ "#Calculation\n",
+ "v=1/((((u2-u1)/R)+(u1/u))/u2)\n",
+ "\n",
+ "#Result\n",
+ "print\"Position of the image is\", round(v,0),\"cm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Position of the image is 259.0 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 94
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 17.14 Page no 932"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "u=-9.0 #cm\n",
+ "y=1\n",
+ "y1=1.5\n",
+ "R=-15.0 #cm\n",
+ "\n",
+ "#Calculation\n",
+ "v=1/(((y-y1)/R)-(y1/-u))\n",
+ "\n",
+ "#Result\n",
+ "print\"The value of distance is\",v,\"cm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of distance is -7.5 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 17.15 Page no 933 "
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "u=-15 #cm\n",
+ "y1=1\n",
+ "y2=1.5\n",
+ "R=-7.5 #cm\n",
+ "\n",
+ "#Calculation\n",
+ "v=1/(((y1-y2)/R)-(y2/-u))\n",
+ "\n",
+ "#Result\n",
+ "print\"Position of the image is\",v,\"cm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Position of the image is -30.0 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 17.16 Page no 933"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "u=-60.0 #cm\n",
+ "R=25.0 #cm\n",
+ "y1=1\n",
+ "y2=1.5\n",
+ "\n",
+ "#Calcution\n",
+ "v=1/((((y2-y1)/R)+(y1/u))/y2)\n",
+ "P=(y2-y1)/(R*10**-2)\n",
+ "\n",
+ "#Result\n",
+ "print\"Position of the image is\", v,\"cm\"\n",
+ "print\"Power of the refracting surface is\", P,\"dioptre\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Position of the image is 450.0 cm\n",
+ "Power of the refracting surface is 2.0 dioptre\n"
+ ]
+ }
+ ],
+ "prompt_number": 21
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 17.17 Page no 934"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "u1=1\n",
+ "u2=1.5\n",
+ "R=1\n",
+ "\n",
+ "#Calculation\n",
+ "x=(u1+u2)/(u2-u1)\n",
+ "\n",
+ "#Result\n",
+ "print\"Distance of the object is\", x,\"R\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Distance of the object is 5.0 R\n"
+ ]
+ }
+ ],
+ "prompt_number": 24
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 17.18 Page no 934"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R=7.5 #cm\n",
+ "u1=1\n",
+ "u2=1.33\n",
+ "\n",
+ "#Calculation\n",
+ "v=1/(((u1-u2)/R))\n",
+ "\n",
+ "#Result\n",
+ "print\"It gets focused at\", round(v,1),\"cm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "It gets focused at -22.7 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 30
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 17.19 Page no 935"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "u1=1\n",
+ "u2=1.5\n",
+ "u=-10\n",
+ "v=-40 #cm\n",
+ "\n",
+ "#Calculation\n",
+ "R=-v*(u2-u1)/(u1+u2)\n",
+ "\n",
+ "#Result\n",
+ "print\"Curvature given to the bounding surface is\", R,\"cm (Convex)\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Curvature given to the bounding surface is 8.0 cm (Convex)\n"
+ ]
+ }
+ ],
+ "prompt_number": 34
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 17.20 Page no 935"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "u1=1\n",
+ "u2=1.5\n",
+ "v=100 #cm\n",
+ "R=20.0 #cm\n",
+ "a=3\n",
+ "b=200.0\n",
+ "\n",
+ "#Calculation\n",
+ "u1=(u2-u1)/R\n",
+ "u2=-1/(u1-(a/b))\n",
+ "d=-u2+R\n",
+ "\n",
+ "#Result\n",
+ "print\"The object distance from the centre of curvature is\", d,\"cm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The object distance from the centre of curvature is 120.0 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 41
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 17.21 Page no 952"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "ug=1.5\n",
+ "R1=50.0 #cm\n",
+ "R2=-50.0 #cm\n",
+ "uw=9/8.0\n",
+ "\n",
+ "#Calculation\n",
+ "f=1/((ug-1)*((1/R1)+(1/R1)))\n",
+ "f1=1/((uw-1)*((1/R1)+(1/R1)))\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Focal length in air is\", f,\"cm\"\n",
+ "print\"(ii) Focal lenth in water is\", f1"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Focal length in air is 50.0 cm\n",
+ "(ii) Focal lenth in water is 200.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 50
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 17.22 Page no 953"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "fa=20 #cm\n",
+ "ug=9/8.0\n",
+ "uw=3/2.0\n",
+ "\n",
+ "#Calculation\n",
+ "a=(uw-1)/(ug-1)\n",
+ "fw=a*fa\n",
+ "f=fw-fa\n",
+ "\n",
+ "#Result\n",
+ "print\"Change in focal length is\", f,\"cm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Change in focal length is 60.0 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 54
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 17.23 Page no 953"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "u=1.56\n",
+ "R1=20.0 #cm\n",
+ "u1=-10.0 #cm\n",
+ "\n",
+ "#Calculation\n",
+ "f=1/((u-1)*(2/R1))\n",
+ "v=1/((1/u1)+(1/f))\n",
+ "\n",
+ "#Result\n",
+ "print\"Position of the image formed is\", round(v,2)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Position of the image formed is -22.73\n"
+ ]
+ }
+ ],
+ "prompt_number": 63
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 17.24 Page no 953"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "u=1.47\n",
+ "\n",
+ "#Calculation\n",
+ "u1=u\n",
+ "\n",
+ "#Result\n",
+ "print\"The liquid is not water because refractive index of water is 1.33\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The liquid is not water because refractive index of water is 1.33\n"
+ ]
+ }
+ ],
+ "prompt_number": 65
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 17.25 Page no 954"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "f=18 #cm\n",
+ "u=1.5\n",
+ "\n",
+ "#Calculation\n",
+ "R=(u-1)*f\n",
+ "\n",
+ "#Result\n",
+ "print\"Radius of the curvature is\", R,\"cm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Radius of the curvature is 9.0 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 68
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 17.26 Page no 954"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "u=-25.0 #cm\n",
+ "f=10.0 #cm\n",
+ "h1=5\n",
+ "\n",
+ "#Calculation\n",
+ "v=1/((1/f)+(1/u))\n",
+ "h2=(v*h1)/u\n",
+ "\n",
+ "#Result\n",
+ "print\"Position of the image is\", round(v,2),\"cm\"\n",
+ "print\"Size of the image is\",round(h2,2),\"cm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Position of the image is 16.67 cm\n",
+ "Size of the image is -3.33 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 78
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 17.27 Page no 954"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "f=-15.0 #cm\n",
+ "v=-10.0 #cm\n",
+ "\n",
+ "#Calculation\n",
+ "u=1/((1/v)-1/f)\n",
+ "\n",
+ "#Result\n",
+ "print\"The object is placed at a distance of\", u,\"cm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The object is placed at a distance of -30.0 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 81
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 17.28 Page no 954"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "v=-20.0 #cm\n",
+ "u=-60.0 #cm\n",
+ "\n",
+ "#Calculation\n",
+ "f=1/((1/v)-(1/u))\n",
+ "\n",
+ "#Result\n",
+ "print\"Focal length of the lens is\", f,\"cm\"\n",
+ "print\"The lens is diverging\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Focal length of the lens is -30.0 cm\n",
+ "The lens is diverging\n"
+ ]
+ }
+ ],
+ "prompt_number": 85
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 17.29 Page no 955"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "u=-10.0 #cm\n",
+ "m=-3.0\n",
+ "\n",
+ "#Calculation\n",
+ "v=m*u\n",
+ "f=1/((1/v)-(1/u))\n",
+ "\n",
+ "#Result\n",
+ "print\"Image formed at\",v,\"cm\"\n",
+ "print\"Focal length is\",f,\"cm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Image formed at 30.0 cm\n",
+ "Focal length is 7.5 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 94
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 17.30 Page no 955"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "P1=6\n",
+ "P2=-2.0\n",
+ "\n",
+ "#Calculation\n",
+ "P=P1+P2\n",
+ "f=1/P\n",
+ "\n",
+ "#Result\n",
+ "print\"Focal length of the combination is\", f*10**2,\"cm\"\n",
+ "print\"Power of the combinationis\",P,\"D\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Focal length of the combination is 25.0 cm\n",
+ "Power of the combinationis 4.0 D\n"
+ ]
+ }
+ ],
+ "prompt_number": 102
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 17.31 Page no 955"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "f1=20.0 #cm\n",
+ "f2=-40.0 #cm\n",
+ "\n",
+ "#Calculation\n",
+ "f=1/((1/f1)+(1/f2))\n",
+ "P=1/f\n",
+ "\n",
+ "#Result\n",
+ "print\"Focal length is\", f,\"cm\"\n",
+ "print\"Power is\",P*10**2,\"D\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Focal length is 40.0 cm\n",
+ "Power is 2.5 D\n"
+ ]
+ }
+ ],
+ "prompt_number": 107
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 17.32 Page no 955"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=2.0\n",
+ "b=1\n",
+ "\n",
+ "#Calculation\n",
+ "u=(b/a)+b\n",
+ "\n",
+ "#Result\n",
+ "print\"Refractive index of the material is\", u"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Refractive index of the material is 1.5\n"
+ ]
+ }
+ ],
+ "prompt_number": 109
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 17.33 Page no 955"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "f=-0.2 #m\n",
+ "v=0.3 #m\n",
+ "\n",
+ "#Calculation\n",
+ "u=1/((1/v)-(1/f))\n",
+ "\n",
+ "#Result\n",
+ "print\"Position of the point is\", u,\"m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Position of the point is 0.12 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 113
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 17.35 Page no 957"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "u1=-30.0 #cm\n",
+ "f1=10.0\n",
+ "u2=10\n",
+ "f2=-10.0\n",
+ "\n",
+ "#calculation\n",
+ "v1=1/((1/u1)+(1/f1))\n",
+ "v2=1/((1/u2)+(1/f2))\n",
+ "v3=-u1\n",
+ "\n",
+ "#Result\n",
+ "print\"Position of the image for first lens is\", v1,\"cm\"\n",
+ "print\"Position of the image for second lens is\", round(v2*10**-2,0),\"cm\"\n",
+ "print\"Position of the image for third lens is\", v3,\"cm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Position of the image for first lens is 15.0 cm\n",
+ "Position of the image for second lens is -0.0 cm\n",
+ "Position of the image for third lens is 30.0 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 127
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter18_1.ipynb b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter18_1.ipynb new file mode 100644 index 00000000..46d58979 --- /dev/null +++ b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter18_1.ipynb @@ -0,0 +1,549 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:b23935cae4f05cd3030e505584dd90917a65a9efe1b245c771989ab0310cdeb5"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 18 Dispersion of light"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 18.1 Page no 986"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "A=60 #Degree\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "a=math.sqrt(2)*math.sin(30*3.14/180.0)\n",
+ "b=math.asin(a)*180/3.14\n",
+ "c=(b*2)-A\n",
+ "i=(A+c)/2.0\n",
+ "r=A/2.0\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Angle of minimum deviation is\", round(c,0),\"Degree\"\n",
+ "print\"(ii) Angle of incidence is\", round(i,0),\"Degree\"\n",
+ "print\"(iii) The angle of refraction is\", r,\"Degree\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Angle of minimum deviation is 30.0 Degree\n",
+ "(ii) Angle of incidence is 45.0 Degree\n",
+ "(iii) The angle of refraction is 30.0 Degree\n"
+ ]
+ }
+ ],
+ "prompt_number": 17
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 18.2 Page no 986"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=51 #Degree\n",
+ "A=60 #Degree\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "b=(A+a)/2.0\n",
+ "c=A/2.0\n",
+ "u=(math.sin(b*3.14/180.0))/(math.sin(c*3.14/180.0))\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) The refracting angle of the prism is\", A,\"Degree\"\n",
+ "print\"(ii) The refractive index of the material is\",round(u,4)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) The refracting angle of the prism is 60 Degree\n",
+ "(ii) The refractive index of the material is 1.6485\n"
+ ]
+ }
+ ],
+ "prompt_number": 30
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 18.3 Page no 987"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "i1=48 #Degree\n",
+ "A=60 #Degree\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "r=A/2.0\n",
+ "u=math.sin(i1*3.14/180.0)/math.sin(r*3.14/180.0)\n",
+ "\n",
+ "#Result\n",
+ "print\"Refractive index of the material is\", round(u,2)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Refractive index of the material is 1.49\n"
+ ]
+ }
+ ],
+ "prompt_number": 35
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 18.4 Page no 987"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=2.0\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "a=math.sqrt(a)/a\n",
+ "i=math.asin(a)*180/3.14\n",
+ "\n",
+ "#Result\n",
+ "print\"Angle of incidence is\", round(i,0),\"Degree\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Angle of incidence is 45.0 Degree\n"
+ ]
+ }
+ ],
+ "prompt_number": 42
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 18.5 Page no 987"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "u=1.5\n",
+ "a=6 #Degree\n",
+ "\n",
+ "#Calculation\n",
+ "A=a/(u-1)\n",
+ "\n",
+ "#Result\n",
+ "print\"Angle of the prism is\", A,\"Degree\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Angle of the prism is 12.0 Degree\n"
+ ]
+ }
+ ],
+ "prompt_number": 45
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 18.6 Page no 987"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "ug=1.5\n",
+ "r1=30 #Degree\n",
+ "ua=1.0\n",
+ "A=60 #Degree\n",
+ "A1=90 #Degree\n",
+ "\n",
+ "#Calculation\n",
+ "sin=(ug*math.sin(r1*3.14/180.0))/ua\n",
+ "i1=math.asin(sin)*180/3.14\n",
+ "a=(2*i1)-A\n",
+ "sin1=1/ug\n",
+ "r1=math.asin(sin1)*180/3.14\n",
+ "r2=A-r1\n",
+ "sin2=(ug*math.sin(r2*3.14/180.0))\n",
+ "i2=math.asin(sin2)*180/3.14\n",
+ "A3=A1+i2-A\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) The angle of incidence for minimum deviation is\", round(i1,0),\"Degree\"\n",
+ "print\"(ii) The angle of minimum deviation is\", round(a,0)\n",
+ "print\"(iii) The angle of emergence of light at maximum deviation is\", round(i2,0),\"Degree\"\n",
+ "print\"(iv) Angle of maximum deviation is\", round(A3,0)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) The angle of incidence for minimum deviation is 49.0 Degree\n",
+ "(ii) The angle of minimum deviation is 37.0\n",
+ "(iii) The angle of emergence of light at maximum deviation is 28.0 Degree\n",
+ "(iv) Angle of maximum deviation is 58.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 68
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 18.7 Page no 991"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "uv=1.68\n",
+ "ur=1.56\n",
+ "A=18 #degree\n",
+ "\n",
+ "#Calculation\n",
+ "A1=A*(uv-ur)\n",
+ "\n",
+ "#Result\n",
+ "print\"Angular dispersion is\", A1,\"Degree\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Angular dispersion is 2.16 Degree\n"
+ ]
+ }
+ ],
+ "prompt_number": 71
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 18.8 Page no 991"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "av=3.32 #Degree\n",
+ "ar=3.22 #Degree\n",
+ "a=3.27 #Degree\n",
+ "\n",
+ "#Calculation\n",
+ "w=(av-ar)/a\n",
+ "\n",
+ "#Result\n",
+ "print\"Dispersive power of the flint glass is\", round(w,4)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Dispersive power of the flint glass is 0.0306\n"
+ ]
+ }
+ ],
+ "prompt_number": 74
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 18.9 Page no 991"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "ur=1.5155\n",
+ "uv=1.5245\n",
+ "\n",
+ "#Calculation\n",
+ "u=(ur+uv)/2.0\n",
+ "w=(uv-ur)/(u-1)\n",
+ "\n",
+ "#Result\n",
+ "print\"Dispersive power of the crown glass is\", round(w,4)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Dispersive power of the crown glass is 0.0173\n"
+ ]
+ }
+ ],
+ "prompt_number": 79
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 18.10 Page no 991"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#given\n",
+ "w=0.031\n",
+ "ur=1.645\n",
+ "ub=1.665\n",
+ "\n",
+ "#Calculation\n",
+ "u=1+((ub-ur))/w\n",
+ "\n",
+ "#Result\n",
+ "print\"Refrective index for yellow colour is\", round(u,3)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Refrective index for yellow colour is 1.645\n"
+ ]
+ }
+ ],
+ "prompt_number": 83
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 18.11 Page no 992"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "A=5 #Degree\n",
+ "uv=1.523\n",
+ "ur=1.515\n",
+ "uv1=1.688\n",
+ "ur1=1.650\n",
+ "\n",
+ "#Calculation\n",
+ "u=(uv+ur)/2.0\n",
+ "u1=(uv1+ur1)/2.0\n",
+ "A1=-((u-1)*A)/(u1-1)\n",
+ "\n",
+ "#Result\n",
+ "print\"Angle of flint line is\",round(A1,2),\"degree\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Angle of flint line is -3.88 degree\n"
+ ]
+ }
+ ],
+ "prompt_number": 91
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 18.12 Page no 992"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "w=0.021\n",
+ "u=1.53\n",
+ "w1=0.045\n",
+ "u1=1.65\n",
+ "A1=4.20 #Degree\n",
+ "\n",
+ "#Calculation\n",
+ "A=-(w1*A1*(u1-1))/(w*(u-1))\n",
+ "\n",
+ "#Result\n",
+ "print\"Angle of the prism is\", round(A,2),\"Degree\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Angle of the prism is -11.04 Degree\n"
+ ]
+ }
+ ],
+ "prompt_number": 95
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 18.13 Page no 992"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "A=72 #Degree\n",
+ "ab=56.4 #Degree\n",
+ "ar=53 #Degree\n",
+ "ay=54.6 #Degree\n",
+ "az=54\n",
+ "A11=60 #Degree\n",
+ "ab1=52.8 \n",
+ "A12=50.6\n",
+ "A13=51.9\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "A1=(A+ay)/2.0\n",
+ "A2=A/2.0\n",
+ "ub=(math.sin(A1*3.14/180.0))/(math.sin(A2*3.14/180.0))\n",
+ "A3=(A+ar)/2.0\n",
+ "ur=(math.sin(A3*3.14/180.0))/(math.sin(A2*3.14/180.0))\n",
+ "A4=(A+az)/2.0\n",
+ "uy=(math.sin(A4*3.14/180.0))/(math.sin(A2*3.14/180.0))\n",
+ "w=(ub-ur)/(uy-1)\n",
+ "\n",
+ "#For flint glass prism\n",
+ "A5=(A11+ab1)/2.0\n",
+ "A51=A11/2.0\n",
+ "ub1=(math.sin(A5*3.14/180.0))/(math.sin(A51*3.14/180.0))\n",
+ "A6=(A11+A12)/2.0\n",
+ "ur1=(math.sin(A6*3.14/180.0))/(math.sin(A51*3.14/180.0))\n",
+ "A7=(A11+A13)/2.0\n",
+ "uy1=(math.sin(A7*3.14/180.0))/(math.sin(A51*3.14/180.0))\n",
+ "w1=(ub1-ur1)/(uy1-1)\n",
+ "w2=w/w1\n",
+ "\n",
+ "#Result\n",
+ "print\"The ratio of dispersive power of crown glass and flint glass prism is\", round(w2,3)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The ratio of dispersive power of crown glass and flint glass prism is 0.64\n"
+ ]
+ }
+ ],
+ "prompt_number": 122
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter19_1.ipynb b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter19_1.ipynb new file mode 100644 index 00000000..3a7eac21 --- /dev/null +++ b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter19_1.ipynb @@ -0,0 +1,841 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:91c393a3f1616e8ab4ec5b337712a2b4f8e8dca0635755aebf9e9a0db573e23b"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 19 Optical instruments"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 19.1 Page no 1013"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "v=-75.0\n",
+ "u=0\n",
+ "\n",
+ "#Calculation\n",
+ "f=v\n",
+ "P=100/f\n",
+ "\n",
+ "#Result\n",
+ "print\"Focal length is\", f,\"cm\"\n",
+ "print\"Power of the lens is\",round(P,2),\"D\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Focal length is -75.0 cm\n",
+ "Power of the lens is -1.33 D\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 19.2 Page no 1014"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "u=-25.0 #cm\n",
+ "v=-150.0 #cm\n",
+ "\n",
+ "#Calculation\n",
+ "f=1/((1/v)-1/u)\n",
+ "P=100/f\n",
+ "\n",
+ "#Result\n",
+ "print\"Focal length of the lens is\", f,\"cm\"\n",
+ "print\"Power of the lens is\",round(P,2),\"D\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Focal length of the lens is 30.0 cm\n",
+ "Power of the lens is 3.33 D\n"
+ ]
+ }
+ ],
+ "prompt_number": 14
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 19.3 Page no 1014"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "u=-25.0 #cm\n",
+ "v=-50.0 #cm\n",
+ "\n",
+ "#Calculation\n",
+ "f=1/((1/v)-1/u)\n",
+ "\n",
+ "#Result\n",
+ "print\"Focal length is\", f,\"cm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Focal length is 50.0 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 17
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 19.4 Page no 1014"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "v=-80.0 #cm\n",
+ "\n",
+ "#Calculation\n",
+ "f=v\n",
+ "P=100/f\n",
+ "\n",
+ "#Result\n",
+ "print\"(a) Power of the lens is\", P,\"D\"\n",
+ "print\"(b) No the corrective lens is concave and it reduces the size of the image. Because it bring the object at the far point of the eye\"\n",
+ "print\"(c) The myopic person may have a normal near point. He must keep the book at a distance greater than 25 cm.\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) Power of the lens is -1.25 D\n",
+ "(b) No the corrective lens is concave and it reduces the size of the image. Because it bring the object at the far point of the eye\n",
+ "(c) The myopic person may have a normal near point. He must keep the book at a distance greater than 25 cm.\n"
+ ]
+ }
+ ],
+ "prompt_number": 22
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 19.5 Page no 1015"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "v=-75.0 #cm\n",
+ "u=-25.0 #cm\n",
+ "\n",
+ "#Calculation\n",
+ "f=1/((1/v)-1/u)\n",
+ "P=100/f\n",
+ "\n",
+ "#Result\n",
+ "print\"(a) Power of the lens is\", round(P,2),\"D\"\n",
+ "print\"(b) The corrective lens produce a virtual imageof an object at 25 cm. The angular size of this image is the same as the object\"\n",
+ "print\"(c) A hypermetropic eye may have normal far point.Hence the person prefers not to use the spectacles for distant object\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) Power of the lens is 2.67 D\n",
+ "(b) The corrective lens produce a virtual imageof an object at 25 cm. The angular size of this image is the same as the object\n",
+ "(c) A hypermetropic eye may have normal far point.Hence the person prefers not to use the spectacles for distant object\n"
+ ]
+ }
+ ],
+ "prompt_number": 29
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 19.6 Pageno 1015"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "P=-0.8 #d\n",
+ "v1=-15.0 #cm \n",
+ "v2=-100.0 #cm\n",
+ "\n",
+ "#Calculation\n",
+ "f=100/P\n",
+ "u1=1/((1/v1)-1/f)\n",
+ "u2=1/((1/v2)-(1/f))\n",
+ "\n",
+ "#Result\n",
+ "print\"The person can see objects lying between\",round(-u1,0),\"cm and\",-u2,\"cm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The person can see objects lying between 17.0 cm and 500.0 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 43
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 19.7 Page no 1016"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "u=-25 #cm\n",
+ "p=3.0\n",
+ "\n",
+ "#Calculation\n",
+ "f=100/p\n",
+ "v=1/((1/f)+1/u)\n",
+ "\n",
+ "#Result\n",
+ "print\"Distance is\",round(v,0),\"m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Distance is -1.0 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 48
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 19.8 Page no 1016"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "u=-25.0 #cm\n",
+ "v=-90.0 #cm\n",
+ "\n",
+ "#calculation\n",
+ "f=1/((1/v)-1/u)\n",
+ "f1=(1/2.0)*10**2\n",
+ "u=1/((1/v)-1/f1)\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) focal length is\",round(f,1),\"cm\"\n",
+ "print\"(ii) Distance is\",round(u,1),\"cm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) focal length is 34.6 cm\n",
+ "(ii) Distance is -32.1 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 56
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 19.9 Page no 1022"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "D=25\n",
+ "f=5.0 #cm\n",
+ "\n",
+ "#calculation\n",
+ "M=1+(D/f)\n",
+ "M1=D/f\n",
+ "\n",
+ "#Result\n",
+ "print\"The magnifying power if the final image is formed at the least distance is\",M\n",
+ "print\"The magnifying power if image is formed at infinity is\",M1"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The magnifying power if the final image is formed at the least distance is 6.0\n",
+ "The magnifying power if image is formed at infinity is 5.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 61
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 19.10 Page no 1023"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "f=4.80 #cm\n",
+ "a=1.20\n",
+ "v=-24.0 #cm\n",
+ "\n",
+ "#Calculation\n",
+ "D=f/(a-1)\n",
+ "u=1/((1/v)-1/f)\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) The least distance of distinct vision is\",D,\"cm\"\n",
+ "print\"(ii) Distance from the lens is\",-u,\"cm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) The least distance of distinct vision is 24.0 cm\n",
+ "(ii) Distance from the lens is 4.0 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 69
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 19.11 Page no 1023"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "v0=15.0 #cm\n",
+ "f0=3.0 #cm\n",
+ "D=25\n",
+ "fe=9\n",
+ "\n",
+ "#Calculation\n",
+ "u0=1/((1/v0)-1/f0)\n",
+ "M=-(v0*D)/(u0*fe)\n",
+ "\n",
+ "#Result\n",
+ "print\"Magnifying power is\", round(M,1)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Magnifying power is 11.1\n"
+ ]
+ }
+ ],
+ "prompt_number": 74
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 19.12 Page no 1024"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "P1=1.5 #D\n",
+ "P2=20.0 #D\n",
+ "u=-25.0 #cm\n",
+ "\n",
+ "#Calculation\n",
+ "f2=100/P2\n",
+ "M=1+(D/f2)\n",
+ "f1=100/P1\n",
+ "v=1/((1/f1)+1/u)\n",
+ "M1=1-(v/f2)\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) The maximum magnifying power together with his glasses\", M\n",
+ "print\"(ii) The maximum magnifying power without glasses\",M1"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) The maximum magnifying power together with his glasses 6.0\n",
+ "(ii) The maximum magnifying power without glasses 9.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 83
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 19.13 Page no 1024"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "l=16\n",
+ "d=-2.5 #cm\n",
+ "f0=0.4 #cm\n",
+ "D=25\n",
+ "\n",
+ "#Calculation\n",
+ "v0=l+d\n",
+ "u0=1/((1/v0)-1/f0)\n",
+ "M=-v0*D/(u0*d)\n",
+ "\n",
+ "#Result\n",
+ "print\"Magnifying power of the microscope is\", M"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Magnifying power of the microscope is -327.5\n"
+ ]
+ }
+ ],
+ "prompt_number": 88
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 19.14 Page no 1025"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "f0=1.0\n",
+ "u0=-1.1 #cm\n",
+ "D=25\n",
+ "fe=5.0\n",
+ "ve=25.0\n",
+ "\n",
+ "#Calculation\n",
+ "v0=1/((1/f0)+1/u0)\n",
+ "d=v0+fe\n",
+ "M=-(v0*D)/(u0*fe)\n",
+ "ue=-1/((1/ve)+1/fe)\n",
+ "D1=v0-ue\n",
+ "M1=-(v0/u0)*(1+(D/fe))\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Distance between the lenses when image is at infinity\", d,\"cm\"\n",
+ "print\"Magnifying power is\",M\n",
+ "print\"(ii) Distance between the lenses when image is at distinct vision\",round(D1,2),\"cm\"\n",
+ "print\"Magnifying Power is\",M1"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Distance between the lenses when image is at infinity 16.0 cm\n",
+ "Magnifying power is 50.0\n",
+ "(ii) Distance between the lenses when image is at distinct vision 15.17 cm\n",
+ "Magnifying Power is 60.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 106
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 19.15 Page no 1032"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "f0=200 #cm\n",
+ "fe=5.0 #cm\n",
+ "D=25.0 #cm\n",
+ "\n",
+ "#Calculation\n",
+ "M=(f0/fe)*(1+(fe/D))\n",
+ "M1=f0/fe\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Magnifying power when image is formed at near point is\", M\n",
+ "print\"(ii) Magnifying power when image is formed at infinity\",M1"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Magnifying power when image is formed at near point is 48.0\n",
+ "(ii) Magnifying power when image is formed at infinity 40.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 110
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 19.16 Page no 1033"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "fe=3\n",
+ "M=4\n",
+ "\n",
+ "#Calculation\n",
+ "f0=fe*M\n",
+ "\n",
+ "#Result\n",
+ "print\"Focal length of the lenses is\" ,f0,\"cm and\",fe,\"cm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Focal length of the lenses is 12 cm and 3 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 112
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 19.17 Page no 1033"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "u0=-200.0 #cm\n",
+ "f0=30.0 #cm\n",
+ "fe=3\n",
+ "\n",
+ "#Calculation\n",
+ "v0=1/((1/f0)+1/u0)\n",
+ "a=v0+fe\n",
+ "\n",
+ "#Result\n",
+ "print\"Separation between the objective and eyepiece is\", round(a,1),\"cm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Separation between the objective and eyepiece is 38.3 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 118
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 19.18 Page no 1033"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "ve=24.0\n",
+ "fe=8.0\n",
+ "f0=250.0\n",
+ "a=10\n",
+ "\n",
+ "#Calculation\n",
+ "ue=1/((1/ve)-(1/fe))\n",
+ "D=f0-ue\n",
+ "d=a/2.0\n",
+ "A=d/f0\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Distance between objective and eyepiece is\", D,\"cm\"\n",
+ "print\"(ii) Angle subtended by the sun at the objective is\",A,\"rad\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Distance between objective and eyepiece is 262.0 cm\n",
+ "(ii) Angle subtended by the sun at the objective is 0.02 rad\n"
+ ]
+ }
+ ],
+ "prompt_number": 126
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 19.19 Page no 1034"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "M=-20\n",
+ "R=-120\n",
+ "\n",
+ "#Calculation\n",
+ "f0=R/2.0\n",
+ "fe=f0/M\n",
+ "\n",
+ "#Result\n",
+ "print\"Focal length of eyepiece is\", fe,\"cm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Focal length of eyepiece is 3.0 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 131
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 19.20 Page no 1034"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "fa=180\n",
+ "f=3.5\n",
+ "fe=5.0\n",
+ "\n",
+ "#Calculation\n",
+ "d=fa+(2*f)+(2*f)+fe\n",
+ "M=-fa/fe\n",
+ "\n",
+ "#Result\n",
+ "print\"Magnifying power of thetelescope is\", M"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Magnifying power of thetelescope is -36.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 135
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 19.21 Page no 1034"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "u0=-200.0 #cm\n",
+ "fa=50.0 #cm\n",
+ "ve=-25.0 #cm\n",
+ "fe=5.0 #cm\n",
+ "\n",
+ "#Calculation\n",
+ "v0=1/((1/fa)+1/u0)\n",
+ "M0=v0/u0\n",
+ "ue=1/((1/ve)-1/fe)\n",
+ "Me=ve/ue\n",
+ "D=v0-ue\n",
+ "M=M0*Me\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Saparation between the objective and eyepiece is\", round(D,2),\"cm\"\n",
+ "print\"(ii) Magnification is\",M"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Saparation between the objective and eyepiece is 70.83 cm\n",
+ "(ii) Magnification is -2.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 147
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter1_1.ipynb b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter1_1.ipynb new file mode 100644 index 00000000..f6180b5a --- /dev/null +++ b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter1_1.ipynb @@ -0,0 +1,543 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:9880f2d8505e271317a099910ead6c2116ce86fa0e83f56feb35ac33a1b96b23"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 1 Electric charge"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.1 Page no 9"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "q=4.5*10**-19 #C\n",
+ "e=1.6*10**-19 #C\n",
+ "\n",
+ "#Calculation\n",
+ "n=q/e\n",
+ "\n",
+ "#Result\n",
+ "print\"n= \",round(n,1),\"This value of charge is not possible\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "n= 2.8 This value of charge is not possible\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.2 Page no 9"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "q=3.2*10**-7 #C\n",
+ "e=1.6*10**-19 #C\n",
+ "\n",
+ "#Calculation\n",
+ "n=q/e\n",
+ "\n",
+ "#Result\n",
+ "print\"The required number of electrons is \",n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The required number of electrons is 2e+12\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.3 Page no 9"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "q=19.2*10**-19\n",
+ "e=1.6*10**-19\n",
+ "me=9*10**-31 #Kg\n",
+ "\n",
+ "#Calculation\n",
+ "n=q/e\n",
+ "M=n*me\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) The value of n=\",n,\"\\n(ii) Charge on silk=\",-q*10**19,\"*10**-19\"\n",
+ "print\"(iii) Mass=\",M,\"Therefore mass transferred is negligibly small\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) The value of n= 12.0 \n",
+ "(ii) Charge on silk= -19.2 *10**-19\n",
+ "(iii) Mass= 1.08e-29 Therefore mass transferred is negligibly small\n"
+ ]
+ }
+ ],
+ "prompt_number": 17
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.4 Page no 9"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=16\n",
+ "n=6.023*10**23 #C\n",
+ "\n",
+ "#Calculation\n",
+ "W=2+a\n",
+ "A=((n*100)/W)*10\n",
+ "\n",
+ "#Result\n",
+ "print\"Total number of electrons in 100 g of water \", round(A,-23)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Total number of electrons in 100 g of water 3.35e+25\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.5 Page no 9"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "n=10**9\n",
+ "e=1.6*10**-19 #C\n",
+ "Q=1\n",
+ "\n",
+ "#Calculation\n",
+ "q=n*e\n",
+ "t=Q/q\n",
+ "\n",
+ "#Result\n",
+ "print (t*10**-9),\"10**9 S\"\n",
+ "print\"Time required is about 198 years\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "6.25 10**9 S\n",
+ "Time required is about 198 years\n"
+ ]
+ }
+ ],
+ "prompt_number": 16
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.6 Page no 13"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "q1=20 #micro C\n",
+ "q2=-5 #micro C\n",
+ "a=9*10**9\n",
+ "r=0.1 \n",
+ "\n",
+ "#Calculation\n",
+ "q=q1+q2\n",
+ "q3=q/2.0\n",
+ "F=(a*q3*q3)/r**2\n",
+ "\n",
+ "#Result\n",
+ "print\"Force is \",round(F*10**-13,3),\"N\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Force is 5.062 N\n"
+ ]
+ }
+ ],
+ "prompt_number": 24
+ },
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Example 1.10 Page no 14"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "m=9*10**9\n",
+ "q=5*10**-6\n",
+ "r=0.1\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "F=(m*q*q)/r**2\n",
+ "C=2*F*math.cos(30)*(180/3.14)\n",
+ "\n",
+ "#Result\n",
+ "print\"Force on each charge is \", round(C,1)*10**-1,\"N\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Force on each charge is 39.79 N\n"
+ ]
+ }
+ ],
+ "prompt_number": 66
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.11 Page no 14"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "m=9*10**9\n",
+ "q=1\n",
+ "r=0.24\n",
+ "A=20\n",
+ "B=12.0\n",
+ "m1=10**-4\n",
+ "g=9.8\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "F=(m*q**2)/r**2\n",
+ "AD=math.sqrt(A**2-B**2)\n",
+ "C=AD/B\n",
+ "F1=(1/C)*m1*g\n",
+ "Q=math.sqrt(F1/F)\n",
+ "\n",
+ "#Result\n",
+ "print\"Charge on each sphere\", round(Q*10**8,1),\"10**-8\",\"C\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Charge on each sphere 6.9 10**-8 C\n"
+ ]
+ }
+ ],
+ "prompt_number": 79
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.12 Page no 15"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "F=3.7*10**-9 #C\n",
+ "q=1.6*10**-19 #c\n",
+ "m=9*10**9\n",
+ "r=5*10**-10\n",
+ "\n",
+ "#Calculation \n",
+ "import math\n",
+ "n=math.sqrt(F*r**2/(m*q**2))\n",
+ "\n",
+ "#Result\n",
+ "print round(n,0),\"electrons are missing from each icon\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "2.0 electrons are missing from each icon\n"
+ ]
+ }
+ ],
+ "prompt_number": 82
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.14 Page no 16"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "e=1.6*10**-19\n",
+ "m=9*10**9\n",
+ "G=6.67*10**-11\n",
+ "me=9.11*10**-31\n",
+ "mp=1.67*10**-27\n",
+ "r=10**-10\n",
+ "\n",
+ "#Calculation\n",
+ "F0=(m*e**2)/(G*me*mp)\n",
+ "F1=(m*e**2)/(G*mp*mp)\n",
+ "F2=m*e**2/r**2\n",
+ "A1=F2/me\n",
+ "A2=F2/mp\n",
+ "\n",
+ "#Result\n",
+ "print\"(a)(i)strength of an electrons and protons\", round(F0*10**-39,1)*10**39\n",
+ "print\" (ii)Strength of two protons \",round(F1*10**-36,1)*10**36\n",
+ "print\"(b) Acceleration of electron is \",round(A1*10**-22,1)*10**22,\"m/s**2\"\n",
+ "print\" Acceleration of proton is \",round(A2*10**-19,1)*10**19,\"m/s*2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a)(i)strength of an electrons and protons 2.3e+39\n",
+ " (ii)Strength of two protons 1.2e+36\n",
+ "(b) Acceleration of electron is 2.5e+22 m/s**2\n",
+ " Acceleration of proton is 1.4e+19 m/s*2\n"
+ ]
+ }
+ ],
+ "prompt_number": 112
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.16 Page no 19"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "m=9*10**9 #C\n",
+ "q1=10*10**-6\n",
+ "q2=5*10**-6\n",
+ "r=0.05\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "F1=m*q1*q2/r**2\n",
+ "F2=m*q1*q2/r**2\n",
+ "F3=math.sqrt(F1**2+F2**2+(2*F1*F2*math.cos(120)*180/3.14))\n",
+ "\n",
+ "#Result\n",
+ "print\"Resultant charge is \", round(F3*10**-1,0),\"N\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Resultant charge is 176.0 N\n"
+ ]
+ }
+ ],
+ "prompt_number": 132
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.17 Page no 20 "
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "m=9*10**9\n",
+ "q1=1.2*10**-8\n",
+ "q2=1\n",
+ "r=0.03\n",
+ "r1=0.04\n",
+ "q3=1.6*10**-8\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "F1=m*q1*q2/r**2\n",
+ "F2=m*q3*q2/r1**2\n",
+ "F3=math.sqrt(F1**2+F2**2)\n",
+ "\n",
+ "#Result\n",
+ "print\"Total force is \", F3*10**-5,\"10**5\",\"N\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Total force is 1.5 10**5 N\n"
+ ]
+ }
+ ],
+ "prompt_number": 149
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.18 Page no 20"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "m=9*10**9\n",
+ "q1=1\n",
+ "q2=100\n",
+ "r=10\n",
+ "q3=75 #C\n",
+ "r1=5\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "F=m*q1*q2/r**2 #along BA\n",
+ "F1=m*q1*q2/r**2 #along AC\n",
+ "F2=m*q3/(math.sqrt(r**2-r1**2)**2)\n",
+ "F3=math.sqrt(F1**2+F2**2)\n",
+ "X=F1/F2\n",
+ "\n",
+ "#Result\n",
+ "print\"Force experienced by 1 C Charge is \",round(F3*10**-9,2),\"N\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Force experienced by 1 C Charge is 12.73 N\n"
+ ]
+ }
+ ],
+ "prompt_number": 168
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter20_1.ipynb b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter20_1.ipynb new file mode 100644 index 00000000..e96a7bf3 --- /dev/null +++ b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter20_1.ipynb @@ -0,0 +1,330 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:49e10d509d6c3c83253662b249f2d9cebaf084cb6d339d2868de883d5e7038f4"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 20 Photometry"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 20.1 Page no 1055"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "E=2.5*10**5 #lm/m**2\n",
+ "r=1.5*10**11 #m\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "l=E*r**2\n",
+ "a=4*math.pi*l\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Luminous intensity is\", l,\"cd\"\n",
+ "print\"(ii) Luminous flux of the sun is\",round(a*10**-28,3)*10**28,\"lm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Luminous intensity is 5.625e+27 cd\n",
+ "(ii) Luminous flux of the sun is 7.069e+28 lm\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 20.2 Page no 1055"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "I2=150\n",
+ "I1=75.0\n",
+ "E1=20\n",
+ "\n",
+ "#Calculation\n",
+ "E2=(I2*E1)/I1\n",
+ "\n",
+ "#Result\n",
+ "print\"Illumination is\", E2,\"lux\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Illumination is 40.0 lux\n"
+ ]
+ }
+ ],
+ "prompt_number": 12
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 20.3 Page no 1056"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "I=35\n",
+ "e=5.0 #lumen/watt\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "a=4*math.pi*I\n",
+ "P=a/e\n",
+ "\n",
+ "#Result\n",
+ "print\"Power of the lamp is\", round(P,0),\"Watt\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Power of the lamp is 88.0 Watt\n"
+ ]
+ }
+ ],
+ "prompt_number": 17
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 20.4 Page no 1056"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=1260\n",
+ "r=8 #m\n",
+ "a1=6\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "I=a/(4.0*math.pi)\n",
+ "Ea=I/r**2\n",
+ "LB=math.sqrt(r**2+a1**2)\n",
+ "cos=r/LB\n",
+ "Eb=(I*cos)/LB**2\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) The illumination at a point immediately below the lamp is\", round(Ea,2),\"lux\"\n",
+ "print\"(ii) The illumination on the working plane is\",round(Eb,1),\"lux\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) The illumination at a point immediately below the lamp is 1.57 lux\n",
+ "(ii) The illumination on the working plane is 0.8 lux\n"
+ ]
+ }
+ ],
+ "prompt_number": 32
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 20.5 Page no 1056"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "r=6.0 #m\n",
+ "I=250 #cd\n",
+ "PQ=8\n",
+ "\n",
+ "#Calculation\n",
+ "Ep=I/r**2\n",
+ "LQ=math.sqrt(r**2+PQ**2)\n",
+ "cos=r/LQ\n",
+ "EQ=(I*cos)/LQ**2\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Illumination at a point P is\", round(Ep,2),\"lux\"\n",
+ "print\"(ii) illumination at a point Q is\",EQ,\"lux\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Illumination at a point P is 6.94 lux\n",
+ "(ii) illumination at a point Q is 1.5 lux\n"
+ ]
+ }
+ ],
+ "prompt_number": 40
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 20.6 Page no 1057"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "t1=2.5 #second\n",
+ "r1=0.5\n",
+ "r2=1\n",
+ "\n",
+ "#Calculation\n",
+ "t2=(t1*r2**2)/r1**2\n",
+ "\n",
+ "#Result\n",
+ "print\"exposure time is\",t2,\"s\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "exposure time is 10.0 s\n"
+ ]
+ }
+ ],
+ "prompt_number": 42
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 20.7 Page no 1058"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "i2=60\n",
+ "r2=105.0\n",
+ "r1=70\n",
+ "\n",
+ "#Calculation\n",
+ "i1=(i2*r1**2)/r2**2\n",
+ "\n",
+ "#Result\n",
+ "print\"The luminous intensity of the first lamp is\",round(i1,2),\"cd\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The luminous intensity of the first lamp is 26.67 cd\n"
+ ]
+ }
+ ],
+ "prompt_number": 45
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 20.8 Page no 1059"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "ra=60\n",
+ "rb=45.0\n",
+ "a=40.0\n",
+ "\n",
+ "#Calculation\n",
+ "ia1=(ra**2)/(rb**2)\n",
+ "ia=(ra**2)/(a**2)\n",
+ "i=ia-ia1\n",
+ "A=(i*100)/ia\n",
+ "\n",
+ "#Result\n",
+ "print\"percentage of light is absorbed by the glass is\",round(A,0),\"%\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "percentage of light is absorbed by the glass is 21.0 %\n"
+ ]
+ }
+ ],
+ "prompt_number": 52
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter21_1.ipynb b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter21_1.ipynb new file mode 100644 index 00000000..5e2ded02 --- /dev/null +++ b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter21_1.ipynb @@ -0,0 +1,383 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:5eb9e6d48b48ecf2d0c9ee8abbe7a462b6b60df5a09da8ebed0ac004de2a0383"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 21 Huygen Principle and interference "
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 21.1 Page no 1090"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Goven\n",
+ "d=5*10**-3 #m\n",
+ "D=1.0 #m\n",
+ "b=0.1092*10**-3\n",
+ "\n",
+ "#Calculation\n",
+ "l=(d*b)/D\n",
+ "\n",
+ "#Result\n",
+ "print\"Wavelength of light used is\", l*10**10,\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Wavelength of light used is 5460.0 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 21.2 Page no 1090"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "l=6200*10**-10 #m\n",
+ "D=0.8\n",
+ "b=2.8*10**-3/4.0\n",
+ "\n",
+ "#Calculation\n",
+ "d=(l*D)/b\n",
+ "\n",
+ "#Result\n",
+ "print\"Separation of the two slit is\", round(d*10**3,1),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Separation of the two slit is 0.7 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 21.3 Page no 1090"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=62\n",
+ "l=5893\n",
+ "l1=4358.0\n",
+ "\n",
+ "#Calculation\n",
+ "n=(a*l)/l1\n",
+ "\n",
+ "#Result\n",
+ "print\"Fringes required is\", round(n,0)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Fringes required is 84.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 16
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 21.4 Page no 1091"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "l=6000*10**-10 #m\n",
+ "D=0.800 #m\n",
+ "d=0.200*10**-3 #m\n",
+ "\n",
+ "#Calculation\n",
+ "x2=(3*l*D)/(2.0*d)\n",
+ "x21=(2*D*l)/d\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Distance of the second dark fringe is\", x2*10**2,\"cm\"\n",
+ "print\"(ii) Distance of the second dark fringe is\", x21*10**2,\"cm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Distance of the second dark fringe is 0.36 cm\n",
+ "(ii) Distance of the second dark fringe is 0.48 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 25
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 21.6 Page no 1091"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Imax=16\n",
+ "Imin=4\n",
+ "\n",
+ "#Calculation\n",
+ "r=Imax/Imin\n",
+ "\n",
+ "#Result\n",
+ "print\"Deduce the ratio of intensity is\", r,\":1\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Deduce the ratio of intensity is 4 :1\n"
+ ]
+ }
+ ],
+ "prompt_number": 29
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 21.7 Page no 1092"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "b=2\n",
+ "u=1.33\n",
+ "\n",
+ "#Calculation\n",
+ "b1=b/u\n",
+ "\n",
+ "#Result\n",
+ "print\"Fringe width is\", round(b1,1),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Fringe width is 1.5 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 33
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 21.8 Page no 1092"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "b2=0.4\n",
+ "b1=0.6\n",
+ "l1=5000\n",
+ "\n",
+ "#Calculation\n",
+ "l2=(b2*2*l1)/b1\n",
+ "\n",
+ "#Result\n",
+ "print\"Wavelength of the light is\", round(l2,0),\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Wavelength of the light is 6667.0 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 37
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 21.9 Page no 1092"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "d=0.125*10**-3 #m\n",
+ "l=4500*10**-10 #m\n",
+ "D=1 #m\n",
+ "\n",
+ "#Calculation\n",
+ "x2=(2*D*l)/d\n",
+ "d1=2*x2\n",
+ "\n",
+ "#Result\n",
+ "print\"Separation between the fringes is\", d1*10**3,\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Separation between the fringes is 14.4 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 42
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 21.10 Page no 1092"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Imax=121\n",
+ "Imin=81.0\n",
+ "\n",
+ "#Calculation\n",
+ "a=Imax/Imin\n",
+ "\n",
+ "#Result\n",
+ "print\"The ratio of intensity at the maxima and minima is\",round(a,2)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The ratio of intensity at the maxima and minima is 1.49\n"
+ ]
+ }
+ ],
+ "prompt_number": 46
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 21.13 Page no 1093"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "l=5.0 #m\n",
+ "d=1 #mm\n",
+ "\n",
+ "#Calculation\n",
+ "a=d/l\n",
+ "\n",
+ "#Result\n",
+ "print\"Width of each slit is\", a,\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Width of each slit is 0.2 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 49
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter22_1.ipynb b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter22_1.ipynb new file mode 100644 index 00000000..e91e3a27 --- /dev/null +++ b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter22_1.ipynb @@ -0,0 +1,900 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:c280b16f38bc8dbf3b2a0607bdd0cfd4670bde51a104a8d547b07a434c49c7f5"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 22 Diffraction and polarisation"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Example 22.1 Page no 1124"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "D=1 #m\n",
+ "l=5*10**-7 #m\n",
+ "d=0.1*10**-3 #m\n",
+ "\n",
+ "#Calculation\n",
+ "W=(2*D*l)/d\n",
+ "\n",
+ "#Result\n",
+ "print\"Width of the central maximum is\", W*10**2,\"cm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Width of the central maximum is 1.0 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 22.2 Page no 1124"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "D=1.60 #m\n",
+ "l=6328*10**-10 #m\n",
+ "w=4.0*10**-3\n",
+ "\n",
+ "#Calculation\n",
+ "d=(2*D*l)/w\n",
+ "\n",
+ "#Result\n",
+ "print\"Width of the slit is\", round(d*10**3,2),\"mm\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Width of the slit is 0.51 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 22.3 Page no 1124"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "l=7500*10**-10\n",
+ "d=1.0*10**-6\n",
+ "c=20\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "a=l/d\n",
+ "b=math.asin(a)*180/3.14\n",
+ "A=2*b\n",
+ "x=c*math.tan(a*3.14/180.0)\n",
+ "w=2*x\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Width of central maximum is\", round(A,0),\"Degree\"\n",
+ "print\"(ii) Width of central maximum is\",round(w*10**2,0),\"cm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Width of central maximum is 97.0 Degree\n",
+ "(ii) Width of central maximum is 52.0 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 27
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 22.4 Page no 1125"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "l=6.3*10**-7 #m\n",
+ "a=3.6 #Degree\n",
+ "n=10\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "d=(n*l)/math.sin(a*3.14/180.0)\n",
+ "\n",
+ "#Result\n",
+ "print\"Slit width is\", round(d*10**3,1),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Slit width is 0.1 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 33
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 22.5 Page no 1125"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "l=5500*10**-10\n",
+ "d=0.01\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "a=l/d\n",
+ "b=math.asin(a)*180/3.14\n",
+ "\n",
+ "#Result\n",
+ "print\"Angular deflection is\", round(b,4),\"Degree\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Angular deflection is 0.0032 Degree\n"
+ ]
+ }
+ ],
+ "prompt_number": 38
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 22.6 Page no 1125"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "lr=660\n",
+ "d=3.0\n",
+ "\n",
+ "#Calculation\n",
+ "l1=(2*lr)/d\n",
+ "\n",
+ "#Result\n",
+ "print\"The value of lambda is\",l1"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of lambda is 440.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 40
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 22.7 Page no 1126"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "l1=5890*10**-10 #m\n",
+ "l2=5896*10**-10\n",
+ "d=2.0*10**-6 #m\n",
+ "D=2 #m\n",
+ "\n",
+ "#Calculation\n",
+ "x=(3*D*(l2-l1))/(2*d)\n",
+ "\n",
+ "#Result\n",
+ "print\"Spacing between the first maxima of two sodium lines is\",x*10**4,\"*10**-4 m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Spacing between the first maxima of two sodium lines is 9.0 *10**-4 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 46
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 22.8 Page no 1126"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "d=3*10**-3 #m\n",
+ "l=500*10**-9 #m\n",
+ "\n",
+ "#Calculation\n",
+ "Z=d**2/l\n",
+ "\n",
+ "#Result\n",
+ "print\"Distance is\",Z,\"m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Distance is 18.0 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 48
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 22.9 Page no 1126"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "d=2*10**-3 #m\n",
+ "l=600*10**-9 #m\n",
+ "\n",
+ "#Calculation\n",
+ "Z=d**2/l\n",
+ "\n",
+ "#Result\n",
+ "print\"Distance is\",round(Z,2),\"m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Distance is 6.67 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 51
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 22.10 Page no 1126"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "d=2*10**-3 #m\n",
+ "l=5000*10**-10\n",
+ "\n",
+ "#Calculation\n",
+ "Z=d**2/l\n",
+ "\n",
+ "#Result\n",
+ "print\"Fresnel Distance is\",Z,\"m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Fresnel Distance is 8.0 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 53
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 22.11 Page no 1129"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "l=5.50*10**-7 #m\n",
+ "D=5.1\n",
+ "\n",
+ "#Calculation\n",
+ "a=(1.22*l)/D\n",
+ "\n",
+ "#Result\n",
+ "print\"Minimum angular separation is\", round(a*10**7,1)*10**-7,\"rad\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Minimum angular separation is 1.3e-07 rad\n"
+ ]
+ }
+ ],
+ "prompt_number": 60
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 22.12 Page no 1130"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "l=6*10**-7 #m\n",
+ "D=0.6\n",
+ "l1=10**10 #m\n",
+ "r=10.0**4*9.46*10**15 #m\n",
+ "\n",
+ "#Calculation\n",
+ "a=(1.22*l)/D\n",
+ "a1=l1/r\n",
+ "\n",
+ "#Result\n",
+ "print round(a1*10**10,2)*10**-10,\"rad\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "1.06e-10 rad\n"
+ ]
+ }
+ ],
+ "prompt_number": 66
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 22.13 Page no 1130"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "l=6000*10**-8\n",
+ "D=254.0\n",
+ "\n",
+ "#Calculation\n",
+ "a=(1.22*l)/D\n",
+ "\n",
+ "#Result\n",
+ "print\"Limt of resolution of a telescope is\",round(a*10**7,1)*10**-7,\"Radian\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Limt of resolution of a telescope is 2.9e-07 Radian\n"
+ ]
+ }
+ ],
+ "prompt_number": 70
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 22.14 Page no 1130"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "D=600.0 #cm\n",
+ "l=5.5*10**-5 #cm\n",
+ "d=3.8*10**10 #cm\n",
+ "\n",
+ "#Calculation\n",
+ "a=(1.22*l)/D\n",
+ "x=d*a\n",
+ "\n",
+ "#Result\n",
+ "print\"Separation of two points is\", round(x,0),\"cm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Separation of two points is 4250.0 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 78
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 22.15 Page no 1130"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "d=10**-4 #cm\n",
+ "l=5.8*10**-5 #cm\n",
+ "\n",
+ "#Calculation\n",
+ "Na=l/(2.0*d)\n",
+ "\n",
+ "#Result\n",
+ "print\"Numerical aperature of a microscope is\", Na"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Numerical aperature of a microscope is 0.29\n"
+ ]
+ }
+ ],
+ "prompt_number": 80
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 22.16 Page no 1131"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "u=1\n",
+ "l=600*10**-9 #,\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "rp=(2*u*math.sin(30*3.14/180.0))/l\n",
+ "\n",
+ "#Result\n",
+ "print\"Resolving power of a microscope is\", round(rp*10**-6,2),\"*10**6\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Resolving power of a microscope is 1.67 *10**6\n"
+ ]
+ }
+ ],
+ "prompt_number": 86
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 22.17 Page no 1133"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "l1=15*10**-10 #m\n",
+ "l=6563*10**-10\n",
+ "c=3*10**8 #m/s\n",
+ "\n",
+ "#Calculation\n",
+ "v=(c*l1)/l\n",
+ "\n",
+ "#Result\n",
+ "print\"Speed of star is\", round(v*10**-5,2),\"*10**5 m/s\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Speed of star is 6.86 *10**5 m/s\n"
+ ]
+ }
+ ],
+ "prompt_number": 91
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 22.18 Page no 1133"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "l1=0.032\n",
+ "l=100.0\n",
+ "c=3*10**8\n",
+ "\n",
+ "#Calculation\n",
+ "v=-(l1*c)/l\n",
+ "\n",
+ "#Result\n",
+ "print\"Velocity of star is\",v*10**-4,\"*10**4 m/s\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Velocity of star is -9.6 *10**4 m/s\n"
+ ]
+ }
+ ],
+ "prompt_number": 95
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 22.21 Page no 1142"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=60 #Degree\n",
+ "a1=90\n",
+ "\n",
+ "import math\n",
+ "A=math.tan(a*3.14/180.0)\n",
+ "ap=a1-a\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Refractive index of the medium is\", round(A,3)\n",
+ "print\"(ii) The refracting angle is\",ap,\"degree\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Refractive index of the medium is 1.73\n",
+ "(ii) The refracting angle is 30 degree\n"
+ ]
+ }
+ ],
+ "prompt_number": 118
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 22.22 Page no 1142"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=1.33\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "ap=math.atan(a)*180/3.14\n",
+ "\n",
+ "#Result\n",
+ "print\"Angle of incidence is\", round(ap,0),\"Degree\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Angle of incidence is 53.0 Degree\n"
+ ]
+ }
+ ],
+ "prompt_number": 122
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 22.23 Page no 1142"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "u=1.33\n",
+ "a=90\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "ap=math.atan(u)*180/3.14\n",
+ "A=a-ap\n",
+ "\n",
+ "#Result\n",
+ "print\"Angle between the sun and the horizon is\", round(A,0),\"Degree\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Angle between the sun and the horizon is 37.0 Degree\n"
+ ]
+ }
+ ],
+ "prompt_number": 127
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 22.24 Page no 1142"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "u=1.5\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "ap=math.atan(u)*180/3.14\n",
+ "r=90-ap\n",
+ "\n",
+ "#Result\n",
+ "print\"Angle of refraction is\", round(r,1),\"Degree\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Angle of refraction is 33.7 Degree\n"
+ ]
+ }
+ ],
+ "prompt_number": 132
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 22.25 Page no 1143"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=30 #Degree\n",
+ "I=3 #A\n",
+ "I0=4.0\n",
+ "I1=1\n",
+ "\n",
+ "#Calculation\n",
+ "a=I/I0\n",
+ "a1=I1/I0\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Fraction of maximum light transferred for 30 degree is\", a\n",
+ "print\"(ii) Fraction of maximum light transferred for 60 degree is\", a1"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Fraction of maximum light transferred for 30 degree is 0.75\n",
+ "(ii) Fraction of maximum light transferred for 60 degree is 0.25\n"
+ ]
+ }
+ ],
+ "prompt_number": 135
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 22.26 Page no 1143"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "ap=60 #Degree\n",
+ "u=3\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "a=1/math.sqrt(u)\n",
+ "C=math.asin(a)*180/3.14\n",
+ "\n",
+ "#Result\n",
+ "print\"Critical angle for this medium is\", round(C,2),\"Degree\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Critical angle for this medium is 35.28 Degree\n"
+ ]
+ }
+ ],
+ "prompt_number": 140
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter23_1.ipynb b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter23_1.ipynb new file mode 100644 index 00000000..7d0fa841 --- /dev/null +++ b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter23_1.ipynb @@ -0,0 +1,893 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:38e55bd383948f67d919af3879ad291116d41c75f201f86aa7c1c2e80cc59941"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Cahpter 23 Dual nature of radiation and matter"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 23.1 Page no 1200"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "h=6.62*10**-34 #J\n",
+ "c=3*10**8 #m/s\n",
+ "l=4.0*10**-7 #m\n",
+ "\n",
+ "#Calculation\n",
+ "E=((h*c)/l)/1.6*10**-19\n",
+ "p=h/l\n",
+ "\n",
+ "#Result\n",
+ "print\"Value of energy is\", round(E*10**38,1),\"ev\"\n",
+ "print\"Momentum of photon is\",p,\"kg m/s\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Value of energy is 3.1 ev\n",
+ "Momentum of photon is 1.655e-27 kg m/s\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 23.2 Page no 1200"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "E=75*1.6*10**-19 #J\n",
+ "h=6.62*10**-34 #J s\n",
+ "\n",
+ "#Calculation\n",
+ "f=E/h\n",
+ "l=(12400/E)*1.6*10**-19\n",
+ "f=c/(l*10**10)\n",
+ "\n",
+ "#Result\n",
+ "print\"Frequency of the photon is\", round(f*10**5,0)*10**15,\"Hz\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Frequency of the photon is 1.8e+16 Hz\n"
+ ]
+ }
+ ],
+ "prompt_number": 31
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 23.3 Page no 1200"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "h=6.62*10**-34 #Js\n",
+ "f=880*10**3 #Hz\n",
+ "E1=10*10**3\n",
+ "\n",
+ "#Calculation\n",
+ "E=h*f\n",
+ "n=E1/E\n",
+ "\n",
+ "#Result\n",
+ "print\"Number of photons emitted per second is\", round(n*10**-31,3)*10**31"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Number of photons emitted per second is 1.717e+31\n"
+ ]
+ }
+ ],
+ "prompt_number": 38
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 23.4 Page no 1200"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "w=1.8\n",
+ "h=6.63*10**-34\n",
+ "l=5000*10**-10\n",
+ "m=9.0*10**-31\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "W=12400/w\n",
+ "h1=(((h*c)/l)-(w*1.6*10**-19))\n",
+ "h2=h1/1.6*10**-19\n",
+ "vmax=math.sqrt((2*h1)/m)\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Threshold wavelength is\",round(W,0),\"A\"\n",
+ "print\"(ii) Maximum K.E of emitted photoelectrons is\", round(h2*10**38,3),\"ev\"\n",
+ "print\"(iii) Maximum velocity is\",round(vmax*10**-5,0),\"*10**5 m/s\"\n",
+ "print\"(iv) If the intensity of light is doubled, K.E of emitted electrons will remain unchanged\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Threshold wavelength is 6889.0 A\n",
+ "(ii) Maximum K.E of emitted photoelectrons is 0.686 ev\n",
+ "(iii) Maximum velocity is 5.0 *10**5 m/s\n",
+ "(iv) If the intensity of light is doubled, K.E of emitted electrons will remain unchanged\n"
+ ]
+ }
+ ],
+ "prompt_number": 65
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 23.5 Page no 1201"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "A=2*10**-4\n",
+ "I=30*10**-2\n",
+ "t=1\n",
+ "E=6.62*10**-19\n",
+ "\n",
+ "#Calculation\n",
+ "n=(I*A)/E\n",
+ "\n",
+ "#Result\n",
+ "print\"Rate at which photons strike the surface is\",round(n*10**-13,2)*10**13,\"photons/s\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Rate at which photons strike the surface is 9.06e+13 photons/s\n"
+ ]
+ }
+ ],
+ "prompt_number": 71
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 23.6 Page no 1201"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "h=6.62*10**-34 #Js\n",
+ "c=3*10**8\n",
+ "l=4500*10**-10 #m\n",
+ "w=2.3\n",
+ "\n",
+ "#Calculation\n",
+ "E=(h*c)/l\n",
+ "E1=(E/1.6*10**-19)*10**38\n",
+ "K=E1-w\n",
+ "f0=(w*1.6*10**-19)/h\n",
+ "p=h/l\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) The energy of photon is\", round(E1,1),\"ev\"\n",
+ "print\"(ii) The maximum kinetic energy of emitted electrons is\",round(K,1),\"ev\"\n",
+ "print\"(iii) Threshold frequency for sodium is\",round(f0*10**-14,1)*10**14,\"Hz\"\n",
+ "print\"(iv) Momentum of a photon is\",round(p*10**27,1)*10**-27,\"Kg m/s\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) The energy of photon is 2.8 ev\n",
+ "(ii) The maximum kinetic energy of emitted electrons is 0.5 ev\n",
+ "(iii) Threshold frequency for sodium is 5.6e+14 Hz\n",
+ "(iv) Momentum of a photon is 1.5e-27 Kg m/s\n"
+ ]
+ }
+ ],
+ "prompt_number": 100
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 23.7 Page no 1202"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "l=36.0*10**-8 #m\n",
+ "w0=2*1.6*10**-19 #J\n",
+ "h=6.62*10**-34 #Js\n",
+ "c=3*10**8\n",
+ "e=1.6*10**-19\n",
+ "m=9.0*10**-31\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "l0=(h*c)/w0\n",
+ "E=(h*c)/l\n",
+ "E1=(E/1.6*10**-19)*10**38\n",
+ "K=E1-2\n",
+ "v0=K\n",
+ "vmax=math.sqrt(e*v0*2/m)\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Threshold wavelength is\",round(l0*10**10,0),\"A\"\n",
+ "print\"(ii) Maximum kinetic energy of emitted photoelectrons is\", round(K,3),\"ev\"\n",
+ "print\"(iii) Stopping potential is\",round(v0,3),\"Volts\"\n",
+ "print\"(iv) Velocity is \",round(vmax*10**-5,2),\"*10**5 m/s\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Threshold wavelength is 6206.0 A\n",
+ "(ii) Maximum kinetic energy of emitted photoelectrons is 1.448 ev\n",
+ "(iii) Stopping potential is 1.448 Volts\n",
+ "(iv) Velocity is 7.18 *10**5 m/s\n"
+ ]
+ }
+ ],
+ "prompt_number": 16
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 23.8 Page no 1202"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "h=6.62*10**-34\n",
+ "c=3*10**8\n",
+ "l0=24.8*10**-8\n",
+ "a=1.2\n",
+ "e=1.6*10**-19\n",
+ "\n",
+ "#Calculation\n",
+ "w0=(h*c)/l0\n",
+ "w01=(w0/1.6*10**-19)*10**38\n",
+ "h1=w01+a\n",
+ "C=h1*e\n",
+ "l=(h*c)/C\n",
+ "\n",
+ "#Result\n",
+ "print\"Wavelength of incident light is\", round(l*10**10,0),\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Wavelength of incident light is 2000.0 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 23.9 Page no 1203"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "v1=16.5\n",
+ "V0=6.6 #V\n",
+ "f0=4.6*10**15 #Hz\n",
+ "f=2.2*10**15 #Hz\n",
+ "\n",
+ "#Calculation\n",
+ "h=(e*(v1-V0))/((f0-f))\n",
+ "\n",
+ "#Result\n",
+ "print\"Planck's constant is\", h"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Planck's constant is 6.6e-34\n"
+ ]
+ }
+ ],
+ "prompt_number": 15
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 23.10 Page no 1203"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "h=6.62*10**-34\n",
+ "f0=44*10**13 #Hz\n",
+ "a=11.5*10**14\n",
+ "b=4.4*10**14\n",
+ "e=1.6*10**-19\n",
+ "\n",
+ "#Calculation\n",
+ "w0=((h*f0)/1.6*10**-19)*10**38\n",
+ "h=3/(a-b)\n",
+ "h1=h*e\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Work function of the material is\", round(w0,2),\"ev\"\n",
+ "print\"(ii) Plank's constant is\", round(h1*10**34,2)*10**-34"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Work function of the material is 1.82 ev\n",
+ "(ii) Plank's constant is 6.76e-34\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 23.11 Page no 1204"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "h=6.6*10**-34\n",
+ "c=3*10**8\n",
+ "l=2000*10**-10\n",
+ "w0=4.2*1.6*10**-19\n",
+ "e=1.6*10**-19\n",
+ "\n",
+ "#Calculation\n",
+ "K=((h*c)/l)-w0\n",
+ "v0=K/e\n",
+ "l1=(h*c)/w0\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Potential difference is\", v0,\"V\"\n",
+ "print\"(ii) Wavelength of incident light is\", round(l1*10**10,0),\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Potential difference is 1.9875 V\n",
+ "(ii) Wavelength of incident light is 2946.0 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 18
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 23.12 Page no 1204"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "h=6.6*10**-34\n",
+ "c=3*10**8\n",
+ "w0=2.39*1.6*10**-19\n",
+ "f1=4000.0 #A\n",
+ "f2=6000 #A\n",
+ "m=9.1*10**-31\n",
+ "e=1.9*10**-19\n",
+ "d=0.1\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "l=(h*c)/w0\n",
+ "K=(12400/f1)-2.39\n",
+ "vmax=math.sqrt((2*K*1.6*10**-19)/m)\n",
+ "B=(m*vmax)/(e*d)\n",
+ "\n",
+ "#Result\n",
+ "print\"Maximum value of B is\", round(B*10**5,2)*10**-5,\"T\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Maximum value of B is 2.39e-05 T\n"
+ ]
+ }
+ ],
+ "prompt_number": 33
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 23.13 Page no 1204"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "w0=4.4\n",
+ "\n",
+ "#Calculation\n",
+ "l=12400/w0\n",
+ "\n",
+ "#Result\n",
+ "print\"Wavelength of visible light is\", round(l,0),\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Wavelength of visible light is 2818.0 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 39
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 23.14 Page no 1205"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "h=6.625*10**-34\n",
+ "c=3*10**8\n",
+ "l=5600*10**-10\n",
+ "a=5\n",
+ "\n",
+ "#Calculation\n",
+ "E=(h*c)/l\n",
+ "n=a/E\n",
+ "\n",
+ "#Result\n",
+ "print\"Number of visible photons emitted per second is\", round(n*10**-19,2)*10**19"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Number of visible photons emitted per second is 1.41e+19\n"
+ ]
+ }
+ ],
+ "prompt_number": 45
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 23.15 Page no 1211"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "v=100\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "l=12.27/math.sqrt(v)\n",
+ "\n",
+ "#Result\n",
+ "print\"Wavelength of an electron is\", l,\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Wavelength of an electron is 1.227 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 49
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 23.16 Page no 1212"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "h=6.62*10**-34\n",
+ "m=9*10**-31\n",
+ "v=10**5\n",
+ "mp=1.67*10**-27\n",
+ "\n",
+ "#Calculation\n",
+ "l=h/(m*v)\n",
+ "lp=h/(mp*v)\n",
+ "\n",
+ "#Result\n",
+ "print\"De-Broglie wavelength of electrons is\", round(l*10**10,1)*10**-10,\"m\"\n",
+ "print\"De-Broglie wavelength of protons is\",round(lp*10**10,4)*10**-10 ,\"m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "De-Broglie wavelength of electrons is 7.36e-09 m\n",
+ "De-Broglie wavelength of protons is 3.96e-12 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 62
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 23.17 Page no 1212"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "E=500*1.6*10**-19\n",
+ "mp=1.67*10**-27\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "l=h/(math.sqrt(2*mp*E))\n",
+ "\n",
+ "#Result\n",
+ "print\"De-Broglie wavelength is\", round(l*10**12,2)*10**-12,\"m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "De-Broglie wavelength is 1.28e-12 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 68
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 23.18 Page no 1212"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "v=150.0\n",
+ "mn=1.675*10**-27 #Kg\n",
+ "En=150*1.6*10**-19\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "le=12.27/math.sqrt(v)\n",
+ "ln=h/math.sqrt(2*mn*En)\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) De-Broglie wavelength of electron is\",round(le,0),\"A\"\n",
+ "print\"(ii) De-Broglie wavelength of neutron is\", round(ln*10**10,4),\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) De-Broglie wavelength of electron is 1.0 A\n",
+ "(ii) De-Broglie wavelength of neutron is 0.0233 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 78
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 23.19 Page no 1213"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "l=2.0*10**-10 #m\n",
+ "h=6.62*10**-34\n",
+ "\n",
+ "#Calculation\n",
+ "p=h/l\n",
+ "\n",
+ "#Result\n",
+ "print\"Momentum of electrons is\", p,\"Kg m/s\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Momentum of electrons is 3.31e-24 Kg m/s\n"
+ ]
+ }
+ ],
+ "prompt_number": 81
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 23.20 Page no 1213"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "l=1.4*10**-10 #m\n",
+ "h=6.63*10**-34\n",
+ "l1=2.0*10**-10\n",
+ "\n",
+ "#Calculation\n",
+ "E=h*c*(1/l-1/l1)\n",
+ "\n",
+ "#Result\n",
+ "print\"Energy of the scattered electron is\", round(E*10**16,2)*10**-16,\"J\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Energy of the scattered electron is 4.26e-16 J\n"
+ ]
+ }
+ ],
+ "prompt_number": 87
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 23.22 Page no 1213"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "me=9.11*10**-31 #Kg\n",
+ "lp=1.813*10**-4\n",
+ "vp=3\n",
+ "\n",
+ "#Calculation\n",
+ "mp=me/(lp*vp)\n",
+ "\n",
+ "#Result\n",
+ "print\"The particle's mass is\", round(mp*10**27,3)*10**-27,\"Kg. The particle is proton\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The particle's mass is 1.675e-27 Kg. The particle is proton\n"
+ ]
+ }
+ ],
+ "prompt_number": 95
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 23.23 Page no 1214"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "l=0.82*10**-10 #m\n",
+ "h=6.6*10**-34\n",
+ "m=9.1*10**-31\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "le=math.sqrt((h*l)/(2*c*m))\n",
+ "\n",
+ "#Result\n",
+ "print\"Wavelength associated with the photoelectrons is\", round(le*10**10,4),\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Wavelength associated with the photoelectrons is 0.0996 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 103
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter24_1.ipynb b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter24_1.ipynb new file mode 100644 index 00000000..b7102dac --- /dev/null +++ b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter24_1.ipynb @@ -0,0 +1,874 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:6d1662d2dadbe072b20c80081401408d705c47c14e10e838032934acc7c20ff4"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 24 Atoms"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 24.1 Page no 1264"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "k=7.68*10**6*1.6*10**-19 #J\n",
+ "e=1.6*10**-19\n",
+ "Z=29\n",
+ "m=9*10**9\n",
+ "\n",
+ "#Calculation\n",
+ "r=(m*2*Z*e**2)/k\n",
+ "\n",
+ "#Result\n",
+ "print\"The distance of the closest approach is\",round(r*10**14,1)*10**-14,\"m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The distance of the closest approach is 1.1e-14 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 24.2 Page no 1265"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=10 #degree\n",
+ "e=1.6*10**-19\n",
+ "Z=79\n",
+ "m=9*10**9\n",
+ "a=5.0*1.6*10**-13\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "b=(Z*e**2*(1/(math.tan(5*3.14/180.0)))*m)/a\n",
+ "\n",
+ "#Result\n",
+ "print\"Impact parameter is\", round(b*10**13,1)*10**-13,\"m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Impact parameter is 2.6e-13 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 20
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 24.3 Page no 1265"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Z=79\n",
+ "m=9*10**9\n",
+ "e=1.6*10**-19\n",
+ "r=4.0*10**-14\n",
+ "\n",
+ "#Calculation\n",
+ "K=(m*2*Z*e**2)/(r*1.6*10**-13)\n",
+ "\n",
+ "#Result\n",
+ "print\"Energy is\", round(K,2),\"Mev\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Energy is 5.69 Mev\n"
+ ]
+ }
+ ],
+ "prompt_number": 25
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 24.4 Page no 1265"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "v=2.1*10**7 #m/s\n",
+ "a=4.8*10**7 #C/Kg\n",
+ "Z=79\n",
+ "e=1.6*10**-19\n",
+ "m=9*10**9\n",
+ "\n",
+ "#Calculation\n",
+ "r0=(2*m*Z*e*a)/v**2\n",
+ "\n",
+ "#Result\n",
+ "print\"Distance of the closest approach is\", round(r0*10**14,1)*10**-14,\"m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Distance of the closest approach is 2.5e-14 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 31
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 24.6 Page no 1266"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Z=79\n",
+ "e=1.6*10**-19 #C\n",
+ "v=1.6*10**-12\n",
+ "m=9*10**9\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "b=(m*Z*e**2*(1/(math.tan(45*3.14/180.0))))/v\n",
+ "\n",
+ "#Result\n",
+ "print\"(a) Scattering angle is 180 degree\"\n",
+ "print\"(b) The value of scattering angle decreases\"\n",
+ "print\"(c) Impact parameter is\", round(b*10**14,1)*10**-14,\"m\"\n",
+ "print\"(d) The scattering of particle takes place due to charge on the nucleus\",\n",
+ "print\"(e) Scattering angle is increase with decrease in impact parameter\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) Scattering angle is 180 degree\n",
+ "(b) The value of scattering angle decreases\n",
+ "(c) Impact parameter is 1.1e-14 m\n",
+ "(d) The scattering of particle takes place due to charge on the nucleus (e) Scattering angle is increase with decrease in impact parameter\n"
+ ]
+ }
+ ],
+ "prompt_number": 47
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 24.7 Page no 1280"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "e=8.854*10**-12\n",
+ "h=6.62*10**-34\n",
+ "m=9*10**-31\n",
+ "e1=1.6*10**-19\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "r1=((e*h**2)/(math.pi*m*e1**2))*10**10\n",
+ "v1=e1**2/(2*e*h)\n",
+ "n=2*r1\n",
+ "\n",
+ "#Result\n",
+ "print\"Radius of first orbit is\", round(r1,2),\"A\"\n",
+ "print\"Velocity of electron is\",round(v1*10**-6,1),\"*10**6 m/s\"\n",
+ "print\"Size of hydrogen atom is\",round(n,2),\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Radius of first orbit is 0.54 A\n",
+ "Velocity of electron is 2.2 *10**6 m/s\n",
+ "Size of hydrogen atom is 1.07 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 62
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 24.8 Page no 1281"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "n=1.0\n",
+ "n1=2.0\n",
+ "n2=3.0\n",
+ "a=0.53*10**-10\n",
+ "Z=3.0\n",
+ "\n",
+ "#Calculation\n",
+ "r1=(a*n)/Z\n",
+ "r2=(a*n1**2)/Z\n",
+ "r3=(a*n2**2)/Z\n",
+ "E1=(-13.6*Z**2)/n**2\n",
+ "E2=(-13.6*Z**2)/n1**2\n",
+ "E3=(-13.6*Z**2)/n2**2\n",
+ "E=E3-E1\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Radii of three lowest allowed orbits is\", round(r1*10**10,2),\"A,\",round(r2*10**10,2),\"A and\",r3*10**10,\"A\"\n",
+ "print\"(ii) Energy of three lowest allowed orbits is\",E1,\"ev,\",E2,\"ev and\",E3,\"ev\"\n",
+ "print\"Energy of the photon is\",E,\"ev\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Radii of three lowest allowed orbits is 0.18 A, 0.71 A and 1.59 A\n",
+ "(ii) Energy of three lowest allowed orbits is -122.4 ev, -30.6 ev and -13.6 ev\n",
+ "Energy of the photon is 108.8 ev\n"
+ ]
+ }
+ ],
+ "prompt_number": 82
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 24.9 Page no 1281"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "n=2.0\n",
+ "n1=3.0\n",
+ "\n",
+ "#Calculation\n",
+ "E2=-13.6/n**2\n",
+ "E3=-13.6/n1**2\n",
+ "\n",
+ "#Result\n",
+ "print\"Energies of two energy level is\",E2,\"ev and\",round(E3,2),\"ev\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Energies of two energy level is -3.4 ev and -1.51 ev\n"
+ ]
+ }
+ ],
+ "prompt_number": 86
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 24.10 Page no 1282"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Rh=1.097*10**7\n",
+ "\n",
+ "#Calculation\n",
+ "l=9/(8.0*Rh)\n",
+ "\n",
+ "#Result\n",
+ "print\"Wavelength of second line is\",round(l*10**10,0),\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Wavelength of second line is 1026.0 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 90
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 24.11 Page no 1282"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Rh=1.097*10**7\n",
+ "\n",
+ "#Calculation\n",
+ "l=4/Rh\n",
+ "\n",
+ "#Result\n",
+ "print\"Shortest wavelength is\",round(l*10**10,0),\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Shortest wavelength is 3646.0 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 95
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 24.12 Page no 1282"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Rh=1.097*10**7\n",
+ "\n",
+ "#Calculation\n",
+ "l=4/(3.0*Rh)\n",
+ "\n",
+ "#Result\n",
+ "print\"Longest wavelength is\",round(l*10**10,0),\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Longest wavelength is 1215.0 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 99
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 24.13 Page no 1282"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "n=1.0\n",
+ "h=6.62*10**-34\n",
+ "c=3*10**8\n",
+ "f=1.6*10**-19\n",
+ "Z=2\n",
+ "\n",
+ "#Calculation\n",
+ "E1=(-13.6*Z**2)/n**2\n",
+ "l=-(h*c)/(E1*f)\n",
+ "\n",
+ "#Result\n",
+ "print\"Minimum wavelength is\", round(l*10**10,0),\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Minimum wavelength is 228.0 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 104
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 24.14 Page no 1283"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "n=1\n",
+ "Z=1.0\n",
+ "a=0.53*10**-10\n",
+ "Z1=4.0\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "rh=(a*n)/Z**2\n",
+ "n1=math.sqrt((a*Z1/rh))\n",
+ "Eh=(-13.6*Z**2)/n**2\n",
+ "Ebe=(-13.6*Z1**2)/n1**2\n",
+ "E=Ebe/Eh\n",
+ "\n",
+ "#Result\n",
+ "print\"Energy of two states is\",E"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Energy of two states is 4.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 112
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 24.15 Page no 1283"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Z=2\n",
+ "e=1.6*10**-19\n",
+ "e1=8.854*10**-12\n",
+ "n=3\n",
+ "h=6.62*10**-34\n",
+ "c=3*10**8\n",
+ "\n",
+ "#Calculation\n",
+ "v=(Z*e**2)/(2*e1*n*h)\n",
+ "a=v/c\n",
+ "\n",
+ "#Result\n",
+ "print\"Speed of the electron is\",round(a,3 )"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Speed of the electron is 0.005\n"
+ ]
+ }
+ ],
+ "prompt_number": 116
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 24.16 Page no 1284"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "r=10**-10\n",
+ "R=10**-15\n",
+ "Rs=7*10**8\n",
+ "\n",
+ "#Calculation\n",
+ "R1=r/R\n",
+ "Re=R1*Rs\n",
+ "\n",
+ "#Result\n",
+ "print\"Radius of the earth's orbit is\",Re,\"m. Thus the earth would be much farther away from the sun\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Radius of the earth's orbit is 7e+13 m. Thus the earth would be much farther away from the sun\n"
+ ]
+ }
+ ],
+ "prompt_number": 118
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 24.17 Page no 1284"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "E=-13.6*1.9*10**-19 #J\n",
+ "m=9*10**9\n",
+ "e=1.6*10**-19\n",
+ "n=1\n",
+ "c=3*10**8\n",
+ "\n",
+ "#Calculation\n",
+ "r=(-e**2*m)/(2.0*E)\n",
+ "v=c/(137*n)\n",
+ "\n",
+ "#Result\n",
+ "print\"Orbital radius is\", round(r*10**11,1)*10**-11,\"m\"\n",
+ "print\"Velocity of the electron is\",round(v*10**-6,1),\"*10**6 m/s\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Orbital radius is 4.5e-11 m\n",
+ "Velocity of the electron is 2.2 *10**6 m/s\n"
+ ]
+ }
+ ],
+ "prompt_number": 131
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 24.18 Page no 1284"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "v=2.2*10**6\n",
+ "r=5.3*10**-11\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "f=v/(2*math.pi*r)\n",
+ "\n",
+ "#Result\n",
+ "print\"Initial frequency of light is\",round(f*10**-15,1)*10**15,\"Hz\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Initial frequency of light is 6.6e+15 Hz\n"
+ ]
+ }
+ ],
+ "prompt_number": 135
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 24.19 Page no 1285"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "m=10 #Kg\n",
+ "T=2*60*60 #S\n",
+ "rn=8*10**6 #m\n",
+ "h=6.62*10**-34\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "vn=(2*math.pi*rn)/T\n",
+ "n=(2*math.pi*rn*vn)/h\n",
+ "\n",
+ "#Result\n",
+ "print\"Quantum number is\",round(n*10**-44,1)*10**45"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Quantum number is 5.3e+45\n"
+ ]
+ }
+ ],
+ "prompt_number": 139
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 24.20 Page no 1285"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "E2=18.70\n",
+ "E1=16.70\n",
+ "h=6.62*10**-34\n",
+ "c=3*10**8\n",
+ "\n",
+ "#Calculation\n",
+ "E=E2-E1\n",
+ "l=(h*c)/(E*1.6*10**-19)\n",
+ "\n",
+ "#Result\n",
+ "print\"Wavelength is\", round(l*10**9,0),\"nm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Wavelength is 621.0 nm\n"
+ ]
+ }
+ ],
+ "prompt_number": 148
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 24.21 Page no 1285"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "n1=2\n",
+ "n2=3\n",
+ "lb=6563\n",
+ "a=20\n",
+ "b=108.0\n",
+ "\n",
+ "#Calculation\n",
+ "l1=(lb*a)/b\n",
+ "\n",
+ "#Result\n",
+ "print\"Wavelength of first member is\",round(l1,0),\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Wavelength of first member is 1215.0 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 151
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 24.22 Page no 1285"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Rh=1.097*10**7 #/m\n",
+ "h=6.63*10**-34\n",
+ "c=3*10**8\n",
+ "n=2.0\n",
+ "n1=4.0\n",
+ "\n",
+ "#Calculation\n",
+ "E=(h*c*Rh*(1/n**2-1/n1**2))/1.6*10**-19\n",
+ "\n",
+ "#Result\n",
+ "print\"Minimum energy is\", round(E*10**38,2),\"ev\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Minimum energy is 2.56 ev\n"
+ ]
+ }
+ ],
+ "prompt_number": 158
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 24.23 Page no 1286"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Rh=1.097*10**7\n",
+ "n2=4.0\n",
+ "n1=3.0\n",
+ "\n",
+ "#Calculation\n",
+ "lm=1/(Rh*(1/n1**2-1/n2**2))\n",
+ "lm1=9/Rh\n",
+ "\n",
+ "#Result\n",
+ "print\"Wavelength is\", round(lm1*10**9,1),\"nm. This wavelength is in infrared part\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Wavelength is 820.4 nm. This wavelength is in infrared part\n"
+ ]
+ }
+ ],
+ "prompt_number": 167
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter25_1.ipynb b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter25_1.ipynb new file mode 100644 index 00000000..95708b68 --- /dev/null +++ b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter25_1.ipynb @@ -0,0 +1,1188 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:efdc68d7aa35d22a94f64e5e8f01516d21c5f3fcea362a5520b7b1d532197f7c"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 25 Nuclei"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 25.1 Page no 1312"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R0=1.2*10**-15 #m\n",
+ "A=208\n",
+ "A1=16\n",
+ "\n",
+ "#calculation\n",
+ "R=R0*A**0.33\n",
+ "R1=R0*A1**0.33\n",
+ "\n",
+ "#Result\n",
+ "print\"Nuclear radius of lead is\", round(R*10**15,1),\"fm\"\n",
+ "print\"Nuclear radius of oxygen is\", round(R1*10**15,0),\"fm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Nuclear radius of lead is 7.0 fm\n",
+ "Nuclear radius of oxygen is 3.0 fm\n"
+ ]
+ }
+ ],
+ "prompt_number": 17
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 25.2 page no1312"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "me=9.1*10**-31\n",
+ "c=3*10**8\n",
+ "e=1.6*10**-19\n",
+ "mp=1.673*10**-27\n",
+ "mn=1.675*10**-27\n",
+ "\n",
+ "#Calculation\n",
+ "E=(me*c**2)/e\n",
+ "E1=(mp*c**2)/e\n",
+ "E2=(mn*c**2)/e\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Equivalent energy of electron is\",round(E*10**-6,2),\"Mev\"\n",
+ "print\"(ii) Equivalent energy of proton is\",round(E1*10**-6,1),\"Mev\"\n",
+ "print\"(iii) Equivalent energy of neutron is\",round(E2*10**-6,1),\"Mev\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Equivalent energy of electron is 0.51 Mev\n",
+ "(ii) Equivalent energy of proton is 941.1 Mev\n",
+ "(iii) Equivalent energy of neutron is 942.2 Mev\n"
+ ]
+ }
+ ],
+ "prompt_number": 27
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 25.3 Page no 1312"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "m=10**-3 #m\n",
+ "c=3*10**8 #m/s\n",
+ "a=3.6*10**6 #J\n",
+ "\n",
+ "#Calculation\n",
+ "E=(m*c**2)/a\n",
+ "\n",
+ "#Result\n",
+ "print E*10**-7,\"*10**7 KWh\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "2.5 *10**7 KWh\n"
+ ]
+ }
+ ],
+ "prompt_number": 30
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 25.4 Page no 1313"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Z=17\n",
+ "A=35\n",
+ "Z1=92\n",
+ "A1=235\n",
+ "Z2=4\n",
+ "A2=9\n",
+ "\n",
+ "#Calculation\n",
+ "n=A-Z\n",
+ "n1=A1-Z1\n",
+ "n2=A2-Z2\n",
+ "\n",
+ "#Calculation\n",
+ "print\"Number of neutron in 17Cl35 is\",n\n",
+ "print\"Number of neutron in 92U235 is\",n1\n",
+ "print\"Number of neutron in 4Be9 is\",n2"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Number of neutron in 17Cl35 is 18\n",
+ "Number of neutron in 92U235 is 143\n",
+ "Number of neutron in 4Be9 is 5\n"
+ ]
+ }
+ ],
+ "prompt_number": 33
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 25.5 Page no 1313"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "A2=235\n",
+ "A1=16.0\n",
+ "R1=3*10**-15 #m\n",
+ "\n",
+ "#Calculation\n",
+ "R=(A2/A1)**0.33\n",
+ "R2=R*R1\n",
+ "\n",
+ "#Result\n",
+ "print\"Nuclear radius is\", round(R2*10**15,3),\"fermi\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Nuclear radius is 7.281 fermi\n"
+ ]
+ }
+ ],
+ "prompt_number": 39
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 25.6 Page no 1313"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "me=55.85\n",
+ "u=1.66*10**-27 #Kg\n",
+ "R=1.2*10**-15 \n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "m=me*u\n",
+ "a=(3*u)/(4.0*math.pi*R**3)\n",
+ "\n",
+ "#Result\n",
+ "print\"Nuclear density is\", round(a*10**-17,2)*10**17,\"Kg/m**3\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Nuclear density is 2.29e+17 Kg/m**3\n"
+ ]
+ }
+ ],
+ "prompt_number": 45
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 25.7 Page no 1317"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "M=4.001509 #a.m.u\n",
+ "N=1.008666\n",
+ "N1=1.007277\n",
+ "a=1.66*10**-27\n",
+ "c=3*10**8\n",
+ "e=1.6*10**-19\n",
+ "n=4.0\n",
+ "\n",
+ "#Calculation\n",
+ "A=2*N1+2*N\n",
+ "M1=A-M\n",
+ "Eb=M1*a*c**2/e\n",
+ "B=Eb/n\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Mass defect is\",M1,\"a.m.u\"\n",
+ "print\"(ii) Binding energy is\",round(Eb*10**-6,1),\"Mev\"\n",
+ "print\" Binding energy per nucleon is\",round(B*10**-6,2),\"Mev\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Mass defect is 0.030377 a.m.u\n",
+ "(ii) Binding energy is 28.4 Mev\n",
+ " Binding energy per nucleon is 7.09 Mev\n"
+ ]
+ }
+ ],
+ "prompt_number": 61
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 25.8 Page no 1317"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "ma=1.00893\n",
+ "m1=1.00813\n",
+ "m2=2.01473\n",
+ "a=931.5\n",
+ "a1=4.00389\n",
+ "\n",
+ "#Calculation\n",
+ "m=ma+m1-m2\n",
+ "Eb=m*a\n",
+ "m3=2*ma+2*m1-a1\n",
+ "Eb1=m3*a\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Binding energy when one neutron and one proton combined together is\", round(Eb,2),\"Mev\"\n",
+ "print\"(ii) Binding energy when two neutrons and two protons are combined is\",round(Eb1,1) ,\"Mev\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Binding energy when one neutron and one proton combined together is 2.17 Mev\n",
+ "(ii) Binding eergy when two neutrons and two protons are combined is 28.2 Mev\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 25.10 Page no 1318"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=1.66*10**-27 #Kg\n",
+ "c=3*10**8\n",
+ "mp=1.00727\n",
+ "mn=1.00866\n",
+ "mo=15.99053\n",
+ "\n",
+ "#Calculation\n",
+ "E=(a*c**2)/1.6*10**-19\n",
+ "m1=8*mp+8*mn-mo\n",
+ "a1=m1*E\n",
+ "\n",
+ "#Result\n",
+ "print\"Energy equivalent of one atomic mass unit is\", round(a1*10**32,1),\"Mev/c**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Energy equivalent of one atomic mass unit is 127.8 Mev/c**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 14
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 25.11 Page no 1318"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "mp=1.007825\n",
+ "mn=1.008665\n",
+ "m=39.962589\n",
+ "a2=931.5\n",
+ "Z=40.0\n",
+ "\n",
+ "#Calculation\n",
+ "E=20*mp+20*mn\n",
+ "m1=E-m\n",
+ "Eb=m1*a2\n",
+ "B=Eb/Z\n",
+ "\n",
+ "#Result\n",
+ "print\"Binding energy per nucleon is\", round(B,3),\"Mev/nucleon\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Binding energy per nucleon is 8.551 Mev/nucleon\n"
+ ]
+ }
+ ],
+ "prompt_number": 21
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 25.12 Page no 1330"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "t=5000 #Days\n",
+ "t1=2000.0\n",
+ "a=0.693 \n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "dt=(a*t)/t1\n",
+ "N=math.log10(dt)\n",
+ "l=a*N/(t1)\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) The fraction remaining after 5000 days is\", round(N,3)\n",
+ "print\"(ii) The activity of sample after 5000 days is\",round(l*10**5,1),\"*10**8 Bq\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) The fraction remaining after 5000 days is 0.239\n",
+ "(ii) The activity of sample after 5000 days is 8.3 *10**8 Bq\n"
+ ]
+ }
+ ],
+ "prompt_number": 42
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 25.13 Page no 1330"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "N=3.67*10**10 #dis/second\n",
+ "r=226.0\n",
+ "A=6.023*10**23\n",
+ "\n",
+ "#Calculation\n",
+ "n=A/r\n",
+ "l=N/n\n",
+ "D=0.693/l\n",
+ "a=D/(3600.0*24.0*365.0)\n",
+ "\n",
+ "#Result\n",
+ "print\" Half life of radium is\",round(a,0),\"years\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " Half life of radium is 1596.0 years\n"
+ ]
+ }
+ ],
+ "prompt_number": 48
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 25.14 page no 1330"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "N0=475\n",
+ "N=270.0\n",
+ "t=5.0\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "a=N0/N\n",
+ "l=math.log(a)/t\n",
+ "T=1/l\n",
+ "T1=0.693/l\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) The decay constant is\",round(l,3),\"/minute\"\n",
+ "print\"(ii) Mean life is\",round(T,2),\"minute\"\n",
+ "print\"(iii) Half life is\",round(T1,2),\"minute\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) The decay constant is 0.113 /minute\n",
+ "(ii) Mean life is 8.85 minute\n",
+ "(iii) Half life is 6.13 minute\n"
+ ]
+ }
+ ],
+ "prompt_number": 55
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 25.15 page no 1331"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "t=1500\n",
+ "N=0.01\n",
+ "N0=0.999\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "T=t*math.log(N)/math.log(0.5)\n",
+ "T1=t*math.log(N0)/math.log(0.5)\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Years will reduce to 1 centigram is\",round(T,1),\"years\"\n",
+ "print\"(ii) Years will lose 1 mg is\",round(T1,2),\"years\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Years will reduce to 1 centigram is 9965.8 years\n",
+ "(ii) Years will lose 1 mg is 2.17 years\n"
+ ]
+ }
+ ],
+ "prompt_number": 61
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 25.16 page no 1331"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=2*10**12\n",
+ "b=9.0*10**12\n",
+ "T=80\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "c=math.log(a/b)\n",
+ "t=-(c*T)/0.693\n",
+ "\n",
+ "#Result\n",
+ "print\"Time required is\",round(t,0),\"second\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Time required is 174.0 second\n"
+ ]
+ }
+ ],
+ "prompt_number": 66
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 25.17 page no 1332"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "T=6.0\n",
+ "A=6.023*10**23\n",
+ "W=99.0\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "l=0.693/T\n",
+ "N0=A*10**-12/W\n",
+ "A0=l*N0\n",
+ "N=N0*(1/math.log10(l))\n",
+ "A1=-(l*N)\n",
+ "\n",
+ "\n",
+ "#Result\n",
+ "print\" Activity in the beginning and after one hour\",round(A1*10**-8,3),\"/h\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " Activity in the beginning and after one hour 7.496 /h\n"
+ ]
+ }
+ ],
+ "prompt_number": 80
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 25.18 page no 1332"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "T=30.0\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "l=0.693/T\n",
+ "T1=1/l\n",
+ "t=math.log(4)/l\n",
+ "t1=math.log(8)/l\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) average life is\",round(l,4),\"/day\"\n",
+ "print\"(ii) The time taken for 3/4 of the original no. to disintegrate is\",round(T1,2),\"days\"\n",
+ "print\"(iii) Time taken is\",round(t,0),\"days\"\n",
+ "print\"(iv) Time taken is\",round(t1,0),\"days\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) average life is 0.0231 /day\n",
+ "(ii) The time taken for 3/4 of the original no. to disintegrate is 43.29 days\n",
+ "(iii) Time taken is 60.0 days\n",
+ "(iv) Time taken is 90.0 days\n"
+ ]
+ }
+ ],
+ "prompt_number": 93
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 25.19 page no 1332"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "l=1620.0\n",
+ "l1=405.0\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "T=(1/l)+(1/l1)\n",
+ "t=math.log(4)/T\n",
+ "\n",
+ "#Result\n",
+ "print\"The time during which three-fourths of a sample will decay is\",round(t,0),\"years\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The time during which three-fourths of a sample will decay is 449.0 years\n"
+ ]
+ }
+ ],
+ "prompt_number": 99
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 25.20 page no 1333"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "C=3.7*10**10 #disintegrations/s\n",
+ "A=6.02*10**23\n",
+ "B=234\n",
+ "\n",
+ "#Calculation\n",
+ "D=(C*B)/A\n",
+ "\n",
+ "#Result \n",
+ "print\"Mass ofuranium atoms disintegrated per second is\",round(D*10**11,3)*10**-11,\"g\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Mass ofuranium atoms disintegrated per second is 1.438e-11 g\n"
+ ]
+ }
+ ],
+ "prompt_number": 105
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 25.21 page no 1333"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "M=0.075 #kg /mol\n",
+ "m=1.2*10**-6 #kg\n",
+ "A=6.0*10**23 #/mol\n",
+ "t=9.6*10**18\n",
+ "N=170\n",
+ "\n",
+ "#Calculation\n",
+ "n=(A*m)/M\n",
+ "l=N/t\n",
+ "T=0.693/l\n",
+ "\n",
+ "#Result\n",
+ "print\"Number of K-40 atoms in the sample is\", n\n",
+ "print\"Half life of K-40 is\", round(T/(24.0*3600.0*365)*10**-9,3),\"*10**9 years\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Number of K-40 atoms in the sample is 9.6e+18\n",
+ "Half life of K-40 is 1.241 *10**9 years\n"
+ ]
+ }
+ ],
+ "prompt_number": 122
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 25.22 Page no 1337"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "mp=232.03714\n",
+ "mn=228.02873\n",
+ "m0=4.002603\n",
+ "a=931.5\n",
+ "A=232.0\n",
+ "e=1.6*10**-19\n",
+ "m=1.66*10**-27\n",
+ "\n",
+ "#Calculation\n",
+ "M=mp-mn-m0\n",
+ "Q=M*a\n",
+ "K=(A-4)*Q/A\n",
+ "S=math.sqrt((2*K*e)/(4.0*m))\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Kinetic energy is\", round(K,1),\"Mev\"\n",
+ "print\"(ii) Speed of particle is\", round(S*10**-4,1),\"*10**7 m/s\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Kinetic energy is 5.3 Mev\n",
+ "(ii) Speed of particle is 1.6 *10**7 m/s\n"
+ ]
+ }
+ ],
+ "prompt_number": 139
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 25.23 Page no 1337"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "b=238\n",
+ "c=206\n",
+ "d=92\n",
+ "e=82\n",
+ "\n",
+ "#Calculation\n",
+ "a=(b-c)/4.0\n",
+ "A=-d+(2*a)+e\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) The emission of alpha particle will reduce the mass number by 4a and charge number by 2a\"\n",
+ "print\"(ii) Number of alpha particle is\", a\n",
+ "print\"Number of beta particle is\",A"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) The emission of alpha particle will reduce the mass number by 4a and charge number by 2a\n",
+ "(ii) Number of alpha particle is 8.0\n",
+ "Number of beta particle is 6.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 145
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 25.24 Page no 1338"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=218\n",
+ "b=84\n",
+ "\n",
+ "#Calculation\n",
+ "A=a-4\n",
+ "Z=b-2\n",
+ "\n",
+ "#Result\n",
+ "print\"Atomic number of new element formed is\", A\n",
+ "print\"Mass number of new element formed is\",Z"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Atomic number of new element formed is 214\n",
+ "Mass number of new element formed is 82\n"
+ ]
+ }
+ ],
+ "prompt_number": 148
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 25.27 Page no 1340"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "mp=10.016125\n",
+ "mn=4.003874\n",
+ "mp1=13.007490\n",
+ "mn1=1.008146\n",
+ "a=931.5\n",
+ "\n",
+ "#Calculation\n",
+ "Mr=mp+mn\n",
+ "Mp=mp1+mn1\n",
+ "Md=Mr-Mp\n",
+ "A=a*Md\n",
+ "\n",
+ "#Result\n",
+ "print\"Energy released in the reaction is\",round(A,3),\"Mev\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Energy released in the reaction is 4.064 Mev\n"
+ ]
+ }
+ ],
+ "prompt_number": 154
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 25.28 Page no 1345"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=10**6 #J/s\n",
+ "E=200*10**6*1.6*10**-19\n",
+ "\n",
+ "#Calculation\n",
+ "N=a/E\n",
+ "\n",
+ "#Result\n",
+ "print\"Number of fission per second is\", round(N*10**-16,2)*10**16"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Number of fission per second is 3.13e+16\n"
+ ]
+ }
+ ],
+ "prompt_number": 159
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 25.29 Page no 1345"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "P=3*10**8 #W\n",
+ "E=200*10**6*1.6*10**-19\n",
+ "a=235\n",
+ "m=6.023*10**23\n",
+ "\n",
+ "#Calculation\n",
+ "E1=P*3600\n",
+ "N=E1/E\n",
+ "M1=(a*N)/m\n",
+ "\n",
+ "#Result\n",
+ "print\"Mass of uranium fissioned per hour is\", round(M1,2),\"g\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Mass of uranium fissioned per hour is 13.17 g\n"
+ ]
+ }
+ ],
+ "prompt_number": 166
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 25.30 Page no 1345"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "m=6.023*10**26\n",
+ "a=235.0\n",
+ "t=30 #Days\n",
+ "E=200*10**6*1.6*10**-19\n",
+ "\n",
+ "#Calculation\n",
+ "N=(2/a)*m\n",
+ "A=N/(t*24*60.0*60.0)\n",
+ "P=E*A\n",
+ "\n",
+ "#Result\n",
+ "print\"Power output is\", round(P*10**-6,1),\"Mev\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Power output is 63.3 Mev\n"
+ ]
+ }
+ ],
+ "prompt_number": 173
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 25.31 Page no 1348"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "m=1.0076\n",
+ "mp=4.0039\n",
+ "a=931.5*10**6 #ev\n",
+ "\n",
+ "#Calculation\n",
+ "Mr=4*m\n",
+ "Md=Mr-mp\n",
+ "E=Md*a*1.6*10**-19\n",
+ "\n",
+ "#Result\n",
+ "print\"Energy released is\", round(E*10**13,2)*10**-13,\"J\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Energy released is 3.95e-12 J\n"
+ ]
+ }
+ ],
+ "prompt_number": 181
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 25.32 Page no 1349"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=6*10**-3 #Kg\n",
+ "c=3*10**8\n",
+ "\n",
+ "#Calculation\n",
+ "E=a*c**2\n",
+ "\n",
+ "#Result\n",
+ "print\"Energy liberated is\", E,\"J\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Energy liberated is 5.4e+14 J\n"
+ ]
+ }
+ ],
+ "prompt_number": 184
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter26_1.ipynb b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter26_1.ipynb new file mode 100644 index 00000000..98e47e10 --- /dev/null +++ b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter26_1.ipynb @@ -0,0 +1,425 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:e41de143240c9df8b907c856d0ba61f830495897881ab9f990dfa099753c5c2e"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 26 Semiconductors"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 26.1 Page no 1414"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=0.47\n",
+ "ue=0.39 #m**2/volt sec\n",
+ "uh=0.19 #m**2/volt sec\n",
+ "e=1.6*10**-19\n",
+ "\n",
+ "#Calculation\n",
+ "a1=1/a\n",
+ "ni=a1/(e*(ue+uh))\n",
+ "\n",
+ "#Result\n",
+ "print\"Intrinsic carrier conceentration is\", round(ni*10**-19,1)*10**19,\"/m**3\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Intrinsic carrier conceentration is 2.3e+19 /m**3\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 26.2 Page no 1414"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=0.01\n",
+ "e=1.6*10**-19\n",
+ "ue=0.39\n",
+ "\n",
+ "#Calculation\n",
+ "a1=1/a\n",
+ "Nd=a1/(e*ue)\n",
+ "\n",
+ "#Result\n",
+ "print\"Donor concentration is\", round(Nd*10**-21,1)*10**21,\"/m**3\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Donor concentration is 1.6e+21 /m**3\n"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 26.3 Page no 1414"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "ni=2.5*10**19 #/m**3\n",
+ "e=1.6*10**19\n",
+ "ue=0.36 #m**2/volt sec\n",
+ "uh=0.17 \n",
+ "\n",
+ "#Calculation\n",
+ "a=ni*e*(ue+uh)\n",
+ "\n",
+ "#Result\n",
+ "print\"Conductivity is\", a*10**-38,\"S/m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Conductivity is 2.12 S/m\n"
+ ]
+ }
+ ],
+ "prompt_number": 17
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 26.4 Page no 1414"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "ne=8*10**13 #/cm**3\n",
+ "nh=5*10**12 #/cm**3\n",
+ "ue=23000 #cm**2/vs\n",
+ "e=1.6*10**-19\n",
+ "uh=100 #cm**2/vs\n",
+ "\n",
+ "#Calculation\n",
+ "a=e*((ne*ue)+(nh*uh))\n",
+ "A1=1/a\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Since electron density is greater than the hole density, the semiconductor is n-type\"\n",
+ "print\"(ii) Resistivity of the sample is\", round(A1,3),\"ohm cm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Since electron density is greater than the hole density, the semiconductor is n-type\n",
+ "(ii) Resistivity of the sample is 3.396 ohm cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 26
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 26.5 Page no 1415"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "ni=1.5*10**16 #/m**3\n",
+ "nh=4.5*10**22 #/m**3\n",
+ "\n",
+ "#Calculation\n",
+ "ne=ni**2/nh\n",
+ "\n",
+ "#Result\n",
+ "print\"ne in the doped semiconductor is\",ne*10**-9,\"*10**9 /m**3\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "ne in the doped semiconductor is 5.0 *10**9 /m**3\n"
+ ]
+ }
+ ],
+ "prompt_number": 29
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 26.6 Page no 1415 "
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "l=5890.0 #A\n",
+ "\n",
+ "#Calculation\n",
+ "E=12400/l\n",
+ "\n",
+ "#Result\n",
+ "print\"Minimum energy is\",round(E,1),\"ev\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Minimum energy is 2.1 ev\n"
+ ]
+ }
+ ],
+ "prompt_number": 32
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 26.7 Page no 1415"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=6*10**19\n",
+ "b=10**-7\n",
+ "\n",
+ "#Calculation\n",
+ "A=a*b\n",
+ "\n",
+ "#Result\n",
+ "print\"Number of holes is\",A"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Number of holes is 6e+12\n"
+ ]
+ }
+ ],
+ "prompt_number": 34
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 26.8 Page no 1415"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "E=0.65\n",
+ "a=10**-10\n",
+ "\n",
+ "#Calculation\n",
+ "l=(12400*a)/E\n",
+ "\n",
+ "#Result\n",
+ "print\"Maximum wavelength of electromagnetic radiation is\",round(l*10**6,1)*10**-6,\"m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Maximum wavelength of electromagnetic radiation is 1.9e-06 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 39
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 26.9 Page no 1416"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=5 #/ohm/cm\n",
+ "ue=3900 #cm**2/vs\n",
+ "e=1.6*10**-19\n",
+ "\n",
+ "#Calculation\n",
+ "Nd=a/(ue*e)\n",
+ "\n",
+ "#Result\n",
+ "print\"Number density of donor atom is\",round(Nd*10**-15,2)*10**15,\"/cm**3\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Number density of donor atom is 8.01e+15 /cm**3\n"
+ ]
+ }
+ ],
+ "prompt_number": 45
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 26.10 Page no 1416"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "ni=1.5*10**16 #/m**3\n",
+ "a=5*10**28\n",
+ "b=10.0**6\n",
+ "\n",
+ "#Calculation\n",
+ "Ne=a/b\n",
+ "nh=ni**2/Ne\n",
+ "\n",
+ "#Result\n",
+ "print\"Number of Electrons is\",Ne,\"/m**3\"\n",
+ "print\"Number of holes is\",nh*10**-9,\"*10**9 /m**3\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Number of Electrons is 5e+22 /m**3\n",
+ "Number of holes is 4.5 *10**9 /m**3\n"
+ ]
+ }
+ ],
+ "prompt_number": 50
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 26.11 Page no 1416"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "d=4.0*10*-8 #m\n",
+ "\n",
+ "#Calculation\n",
+ "a=2/1.6*10**-19\n",
+ "E=-a/d\n",
+ "\n",
+ "#Result\n",
+ "print\"Electric field is\", round(E*10**22,0),\"*10**7 V/m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Electric field is 4.0 *10**7 V/m\n"
+ ]
+ }
+ ],
+ "prompt_number": 61
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter27_1.ipynb b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter27_1.ipynb new file mode 100644 index 00000000..01e54494 --- /dev/null +++ b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter27_1.ipynb @@ -0,0 +1,921 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:eeaae422be8c264750ed4950e51451be86906b82350f05dcb46e0f610199fb25"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 27 Semiconductor devices"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 27.1 Page no 1446"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "E=1.5 #V\n",
+ "Vd=0.5 #V\n",
+ "P=0.1 #W\n",
+ "\n",
+ "#Calculation\n",
+ "Imax=P/Vd\n",
+ "V=E-Vd\n",
+ "R1=V/Imax\n",
+ "\n",
+ "#Result\n",
+ "print\"Value of resistance is\",R1,\"ohm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Value of resistance is 5.0 ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 27.2 Page no 1446"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "V=2 #V\n",
+ "R=10.0 #ohm\n",
+ "R1=20.0\n",
+ "\n",
+ "#Calculation\n",
+ "I=V/R\n",
+ "I1=V/R1\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Current drawn from battery is\", I,\"A\"\n",
+ "print\"(ii) Current drawn from point B is\",I1,\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Current drawn from battery is 0.2 A\n",
+ "(ii) Current drawn from point B is 0.1 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 27.3 Page no 1446"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#given\n",
+ "Vl=15 #V\n",
+ "Rl=2.0*10**3\n",
+ "Iz=10 #mA\n",
+ "\n",
+ "#Calculation\n",
+ "Il=(Vl/Rl)*10**3\n",
+ "Ir=Iz+Il\n",
+ "Vr=Ir*10**-2*R1\n",
+ "V=Vr+Vl\n",
+ "\n",
+ "#Result\n",
+ "print\"Voltage is\", V,\"V\"\n",
+ "print\"Zener rating required is\",Ir,\"mA\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Voltage is 18.5 V\n",
+ "Zener rating required is 17.5 mA\n"
+ ]
+ }
+ ],
+ "prompt_number": 27
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 27.4 Page no 1447"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "N=10.0\n",
+ "V=230 #V\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "Vrpm=math.sqrt(2)*V\n",
+ "Vsm=Vrpm/N\n",
+ "Vdc=Vsm/math.pi\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) The output dc voltage is\", round(Vdc,2),\"V\"\n",
+ "print\"(ii) Peak inverse voltage is\",round(Vsm,2),\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) The output dc voltage is 10.35 V\n",
+ "(ii) Peak inverse voltage is 32.53 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 36
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 27.5 Page no 1447"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Vm=50 #V\n",
+ "rf=20.0\n",
+ "Rl=800 #ohm\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "Im=(Vm/(rf+Rl))*10**3\n",
+ "Idc=Im/math.pi\n",
+ "Irms=Im/2.0\n",
+ "P=(Irms/1000.0)**2*(rf+Rl)\n",
+ "P1=(Idc/1000.0)**2*Rl\n",
+ "V=Idc*Rl*10**-3\n",
+ "A=P1*100/P\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Im=\",round(Im,0),\"mA \\nIdc=\",round(Idc,1),\"mA \\nIrms=\",round(Irms,1),\"mA\"\n",
+ "print\"(ii) a.c power input is\",round(P,3),\"watt \\nd.c. power is\",round(P1,3),\"watt\"\n",
+ "print \"(iii) d.c. output voltage is\",round(V,2),\"Volts\"\n",
+ "print\"(iv) Efficiency of rectification is\", round(A,1),\"%\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Im= 61.0 mA \n",
+ "Idc= 19.4 mA \n",
+ "Irms= 30.5 mA\n",
+ "(ii) a.c power input is 0.762 watt \n",
+ "d.c. power is 0.301 watt\n",
+ "(iii) d.c. output voltage is 15.53 Volts\n",
+ "(iv) Efficiency of rectification is 39.5 %\n"
+ ]
+ }
+ ],
+ "prompt_number": 67
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 27.6 Page no 1448"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "rf=20 #ohm\n",
+ "Rl=980\n",
+ "V=50 #v\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "Vm=V*math.sqrt(2)\n",
+ "Im=(Vm/(rf+Rl))*10**3\n",
+ "Idc=(2*Im)/(math.pi)\n",
+ "Irms=Im/math.sqrt(2)\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) load current is\",round(Im,1),\"mA\"\n",
+ "print\"(ii) Mean load currant is\",round(Idc,0),\"mA\"\n",
+ "print\"(iii) R.M.S value of load current is\",Irms,\"mA\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) load current is 70.7 mA\n",
+ "(ii) Mean load currant is 45.0 mA\n",
+ "(iii) R.M.S value of load current is 50.0 mA\n"
+ ]
+ }
+ ],
+ "prompt_number": 85
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 27.7 page no 1448"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "N=5.0\n",
+ "A=230 #V\n",
+ "B=2\n",
+ "Rl=100\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "V1=A/N\n",
+ "V2=V1*math.sqrt(2)\n",
+ "Vm=V2/B\n",
+ "Idc=2*Vm/(math.pi*Rl)\n",
+ "Vdc=Idc*Rl\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) d.c voltage output is\",round(Vdc,1),\"V\"\n",
+ "print\"(ii) peak inverse voltage is\",round(V2,0),\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) d.c voltage output is 20.7 V\n",
+ "(ii) peak inverse voltage is 65.0 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 122
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 27.8 page no 1448"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Il=4.0 #mA\n",
+ "Vz=6 #V\n",
+ "E=10.0 #V\n",
+ "\n",
+ "#Calculation\n",
+ "Lz=5*Il\n",
+ "L=Il+Lz\n",
+ "Rs=E-Vz\n",
+ "Rs1=Rs/(L*10**-3)\n",
+ "\n",
+ "#Result\n",
+ "print\"The value of series resister Rs\",round(Rs1,0),\"ohm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of series resister Rs 167.0 ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 123
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 27.9 page no 1449"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Vf=0.3 #V\n",
+ "If=4.3*10**-3 #A\n",
+ "Vc=0.35\n",
+ "Va=0.25\n",
+ "Ic=6*10**-3\n",
+ "Ia=3*10**-3\n",
+ "\n",
+ "#Calculation\n",
+ "Rdc=Vf/If\n",
+ "Vf1=Vc-Va\n",
+ "If1=Ic-Ia\n",
+ "Rac=Vf1/If1\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) D.C. resistance is\",round(Rdc,2),\"ohm\"\n",
+ "print\"(ii) A.C. resistance is\",round(Rac,2),\"ohm\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) D.C. resistance is 69.77 ohm\n",
+ "(ii) A.C. resistance is 33.33 ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 139
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 27.10 page no 1462"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "A=0.9\n",
+ "Ie=1 #mA\n",
+ "\n",
+ "#Calculation\n",
+ "Ic=A*Ie\n",
+ "Ib=Ie-Ic\n",
+ "\n",
+ "#Result\n",
+ "print\"Base current is\",Ib,\"mA\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Base current is 0.1 mA\n"
+ ]
+ }
+ ],
+ "prompt_number": 144
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 27.11 page no 1462"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "B=50\n",
+ "Ib=0.02 #mA\n",
+ "\n",
+ "#Calculation\n",
+ "Ic=B*Ib\n",
+ "Ie=Ib+Ic\n",
+ "\n",
+ "#Result\n",
+ "print\"Ie =\",Ie,\"mA\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Ie = 1.02 mA\n"
+ ]
+ }
+ ],
+ "prompt_number": 147
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 27.12 page no 1462"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "B=49\n",
+ "Ie=12 #mA\n",
+ "Ib=240 #microA\n",
+ "\n",
+ "#Calculation\n",
+ "A=(B/1+B)*10**-2\n",
+ "Ic=A*Ie\n",
+ "Ic1=B*Ib\n",
+ "\n",
+ "#Result\n",
+ "print\" The value of Ic using A is\",Ic,\"mA\"\n",
+ "print\" The value of Ic using B is\",Ic1*10**-3,\"mA\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " The value of Ic using A is 11.76 mA\n",
+ " The value of Ic using B is 11.76 mA\n"
+ ]
+ }
+ ],
+ "prompt_number": 165
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 27.13 page no 1463"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "B=45.0\n",
+ "Ic=1 #V\n",
+ "\n",
+ "#Calculation\n",
+ "Ib=Ic/B\n",
+ "\n",
+ "#Result\n",
+ "print\" The base current for common emitter connection is\",round(Ib,3),\"mA\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " The base current for common emitter connection is 0.022 mA\n"
+ ]
+ }
+ ],
+ "prompt_number": 169
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 27.14 page no 1463"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Vcc=8 #V\n",
+ "V=0.5 #V\n",
+ "Rc=800.0 #ohm\n",
+ "a=0.96\n",
+ "\n",
+ "#Calculation\n",
+ "Vce=Vcc-V\n",
+ "Ic=V/Rc*10**3\n",
+ "B=a/(1-a)\n",
+ "Ib=Ic/B\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Collector-emitter voltage is\",Vce,\"V\"\n",
+ "print\"(ii) Base current is\",round(Ib,3),\"mA\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Collector-emitter voltage is 7.5 V\n",
+ "(ii) Base current is 0.026 mA\n"
+ ]
+ }
+ ],
+ "prompt_number": 184
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 27.15 page no 1463"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=10\n",
+ "b=2\n",
+ "c=3\n",
+ "\n",
+ "#Calculation\n",
+ "Vce=a-b\n",
+ "Ic=c-b\n",
+ "Ro=Vce/Ic\n",
+ "\n",
+ "#Result\n",
+ "print\"The output resistance is\",Ro,\"k ohm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The output resistance is 8 k ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 188
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 27.16 page no 1464"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Ic=4.0 #mA\n",
+ "Ib=30 #micro A\n",
+ "Ib1=20 #micro A\n",
+ "Vce=10 #V\n",
+ "c=4.5 #mA\n",
+ "d=3.0 #mA\n",
+ "\n",
+ "#Calculation\n",
+ "Ib2=Ib-Ib1\n",
+ "Ic1=c-d\n",
+ "Bac=Ic1/Ib2*10**3\n",
+ "Bdc=c/Ib*10**3\n",
+ "\n",
+ "#Result \n",
+ "print\"The value of Bac of the transister is\",Bdc\n",
+ "print\"The value of Bdc of the transister is\",Bdc"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of Bac of the transister is 150.0\n",
+ "The value of Bdc of the transister is 150.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 249
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 27.17 page no 1464"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Ri=665.0 #ohm\n",
+ "Ib=15.0 #micro A\n",
+ "Ic=2 #mA\n",
+ "Ro=5*10**3 #ohm\n",
+ "\n",
+ "#Calculation\n",
+ "Bac=Ic/Ib*10**3\n",
+ "Av=Bac*(Ro/Ri)\n",
+ "\n",
+ "#Result\n",
+ "print\" The voltage gain is\", round(Av,0)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " The voltage gain is 1003.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 240
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 27.18 page no 1464"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given \n",
+ "Vbb=2.0 #v\n",
+ "Rc=2000 #ohm\n",
+ "B=100\n",
+ "Vbe=0.6 #V\n",
+ "\n",
+ "#Calculation\n",
+ "Ic=Vbb/Rc*10**3\n",
+ "Ib=Ic/B\n",
+ "Ib1=10*Ib\n",
+ "Rb=(Vbb-Vbe)/Ib\n",
+ "Ic=B*Ib1\n",
+ "\n",
+ "#Result \n",
+ "print\"d.c. collector current is\",Ic,\"mA\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "d.c. collector current is 10.0 mA\n"
+ ]
+ }
+ ],
+ "prompt_number": 236
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 27.19 page no 1465"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "N=10**10\n",
+ "e=1.6*10**-19\n",
+ "t=10**-6\n",
+ "\n",
+ "#Calculation\n",
+ "Ie=(N*e)/t*10**3\n",
+ "Ib=(2/100.0)*Ie\n",
+ "Ic=Ie-Ib\n",
+ "c=Ic/Ie\n",
+ "B=Ic/Ib\n",
+ "#Result\n",
+ "print\"The current transfer ratio\",c\n",
+ "print\"current amplification factor is\",B"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The current transfer ratio 0.98\n",
+ "current amplification factor is 49.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 257
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 27.20 page no 1465"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=200\n",
+ "b=50\n",
+ "c=17\n",
+ "d=5\n",
+ "e=4000\n",
+ "\n",
+ "#Calculation\n",
+ "Ib=(a-b)*10**-3\n",
+ "Ic=c-d\n",
+ "B=Ic/Ib\n",
+ "D=e/B\n",
+ "Ap=B**2*D\n",
+ "\n",
+ "#Result\n",
+ "print\" The value of current gain is\",B\n",
+ "print\" The value of resistance gain is\",D \n",
+ "print\" The value of power gain is\",Ap*10**-5,\"*10**5\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " The value of current gain is 80.0\n",
+ " The value of resistance gain is 50.0\n",
+ " The value of power gain is 3.2 *10**5\n"
+ ]
+ }
+ ],
+ "prompt_number": 279
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 27.21 page no 1469"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "L1=58.6*10**-6 #H\n",
+ "C1=300.0*10**-12 #F\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "f=1/((2.0*math.pi)*math.sqrt(L1*C1))\n",
+ "\n",
+ "#Result\n",
+ "print\"Frequency of oscillation is\", round(f*10**-3,0),\"KHz\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Frequency of oscillation is 1200.0 KHz\n"
+ ]
+ }
+ ],
+ "prompt_number": 294
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 27.22 Page no 1469"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Vbe=0.8 #V\n",
+ "Vcc=5 #V\n",
+ "Rc=1 #K ohm\n",
+ "b=250.0\n",
+ "Rb=100 #K ohm\n",
+ "\n",
+ "#Calculation\n",
+ "Ic=Vcc/Rc\n",
+ "Ib=(Ic/b)*10**3\n",
+ "Vi=(Ib*Rb)+Vbe\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) The minimum base current is\",Ib,\"micro A\"\n",
+ "print\"(ii) The input voltage is\",round(Vi*10**-3,0),\"V\"\n",
+ "print\"(iii) Between 0 V and 0.6 V,the transistor will switched off. Between 2.8 V and 5.0 V the transistor will switched on\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) The minimum base current is 20.0 micro A\n",
+ "(ii) The input voltage is 2.0 V\n",
+ "(iii) Between 0 V and 0.6 V,the transistor will switched off. Between 2.8 V and 5.0 V the transistor will switched on\n"
+ ]
+ }
+ ],
+ "prompt_number": 309
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter2_1.ipynb b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter2_1.ipynb new file mode 100644 index 00000000..e9ce1a12 --- /dev/null +++ b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter2_1.ipynb @@ -0,0 +1,682 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:ef2be2de13061356088f1dea63be1d10f7d8030c93430c7778acfc0d827cf022"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 2 Electric field"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.1 Page no 47"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "V=1000\n",
+ "d=10.0*10**-3\n",
+ "m=4.8*10**-15\n",
+ "g=10\n",
+ "e=1.6*10**-19\n",
+ "\n",
+ "#Calculation\n",
+ "E=V/d\n",
+ "q=m*g/E\n",
+ "n=q/e\n",
+ "\n",
+ "#Result\n",
+ "print\"The number of electrons on the drop \", n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The number of electrons on the drop 3.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2 Page no 47"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "e=1.6*10**-19\n",
+ "E=3*10**4\n",
+ "m=9.0*10**-31\n",
+ "y1=4*10**-2\n",
+ "m2=1.67*10**-27\n",
+ "\n",
+ "#Calculation\n",
+ "a=e*E/m\n",
+ "t=math.sqrt((2*y1)/a)\n",
+ "a2=e*E/m2\n",
+ "t2=math.sqrt((2*y1)/a2)\n",
+ "\n",
+ "#Result\n",
+ "print\"Time t1=\", round(t*10**9,1)*10**-9,\"S\",\"\\nTime t2=\",round(t2*10**7,2)*10**-7,\"S\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Time t1= 3.9e-09 S \n",
+ "Time t2= 1.67e-07 S\n"
+ ]
+ }
+ ],
+ "prompt_number": 21
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.3 Page no 48"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "r=1 #m\n",
+ "m=9*10**9\n",
+ "q=500*10**-6\n",
+ "r1=0.3 #m\n",
+ "\n",
+ "#Calculation\n",
+ "E=m*q/r**2\n",
+ "E2=m*q/r1**2\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Electric field intensity from the centre of the sphere \",E*10**-6,\"10**6\",\"N/C\"\n",
+ "print\"(ii) Electric field intensity at the surface of the sphere is \",E2*10**-7,\"10**7 N/C\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Electric field intensity from the centre of the sphere 4.5 10**6 N/C\n",
+ "(ii) Electric field intensity at the surface of the sphere is 5.0 10**7 N/C\n"
+ ]
+ }
+ ],
+ "prompt_number": 37
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.4 Page no 48"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "q=2*10**-8\n",
+ "E=2*10**4\n",
+ "m=80*10**-6\n",
+ "g=9.8\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "a=q*E/(m*g)\n",
+ "b=math.atan(a)*180/3.14\n",
+ "T=(q*E/(math.sin(b*3.14/180.0)))*10**-4\n",
+ "\n",
+ "#Result\n",
+ "print\"The angle is \", round(b,0),\"degree\"\n",
+ "print\"Tension in the thread of the pendulum is \", round(T*10**8,2),\"*10**-4 N\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The angle is 27.0 degree\n",
+ "Tension in the thread of the pendulum is 8.8 *10**-4 N\n"
+ ]
+ }
+ ],
+ "prompt_number": 33
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.5 Page no 48"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "m=9*10**9\n",
+ "r=0.707\n",
+ "q=5*10**-6\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "E=m*q/r**2 #along AO\n",
+ "E2=m*q/r**2 #along BO\n",
+ "E3=m*q/r**2 #along OD\n",
+ "E11=E+E2\n",
+ "E12=E2+E3\n",
+ "I=(2*E11*r)*10**-4\n",
+ "\n",
+ "#Result\n",
+ "print\"Electric field at the centre of the sphere is \",round(I,2),\"*10**4 N/C\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Electric field at the centre of the sphere is 25.46 *10**4 N/C\n"
+ ]
+ }
+ ],
+ "prompt_number": 53
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.6 Page no 49"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "q=5*10**-9\n",
+ "x=0.15 #m\n",
+ "r=0.1 #m\n",
+ "a=9*10**9\n",
+ "\n",
+ "#Calculation\n",
+ "E=(a*q*x)/((r**2+x**2))**1.5\n",
+ "\n",
+ "#Result\n",
+ "print\"Intensity of the electric field is \", round(E,0),\"N/C\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Intensity of the electric field is 1152.0 N/C\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.7 Page no 49"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "m=10**-3\n",
+ "F=1\n",
+ "v0=20\n",
+ "v=0\n",
+ "\n",
+ "#Calculation\n",
+ "a=-F/m\n",
+ "s=v**2-v0**2/(2.0*a)\n",
+ "\n",
+ "#Result\n",
+ "print\"The distance is \", s,\"m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The distance is 0.2 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.9 Page no 50"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "m=9*10**9\n",
+ "q1=1/3.0*10**-7\n",
+ "r=5*10**-2\n",
+ "F=58.8*10**-3\n",
+ "\n",
+ "#Calculation\n",
+ "q2=F*r**2/(q1*m)\n",
+ "\n",
+ "#Result\n",
+ "print\"Charge is \", q2,\"C\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Charge is 4.9e-07 C\n"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.10 Page no 61"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "E=4*10**5 #V/m\n",
+ "q=3.2*10**-19\n",
+ "a=2.4*10**-10\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "p=q*a\n",
+ "W=p*E*(1-(math.cos(180*180/3.14)))\n",
+ "\n",
+ "#Result\n",
+ "print\"Work done is \", W"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Work done is 2.79670959474e-23\n"
+ ]
+ }
+ ],
+ "prompt_number": 33
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.11 Page no 61"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "q=16*10**-19\n",
+ "a=3.9*10**-12\n",
+ "E=10**5\n",
+ "\n",
+ "#Calculation\n",
+ "p=q*a\n",
+ "U=-p*E\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) The electric dipole moment \", p,\"Cm\"\n",
+ "print\"(ii) Potential energy of dipole in the stable equilibrium position \",U,\"J\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) The electric dipole moment 6.24e-30 Cm\n",
+ "(ii) Potential energy of dipole in the stable equilibrium position -6.24e-25 J\n"
+ ]
+ }
+ ],
+ "prompt_number": 45
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.12 Page no 61"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "q=20*10**-6\n",
+ "a=10**-2\n",
+ "m=9*10**9\n",
+ "r=0.1\n",
+ "\n",
+ "#Calculation\n",
+ "p=q*a\n",
+ "E=m*2*p/r**3\n",
+ "\n",
+ "#Result\n",
+ "print\"Electric field intensity is \", E*10**-5,\"*10**5 N/C\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Electric field intensity is 36.0 *10**5 N/C\n"
+ ]
+ }
+ ],
+ "prompt_number": 51
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.13 Page no 62"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "E=4*10**5\n",
+ "q=1*10**-6\n",
+ "a=3*10**-2\n",
+ "\n",
+ "#Calculation\n",
+ "t=q*a*E\n",
+ "\n",
+ "#Result\n",
+ "print\"Maximum torque on the dipole is \", t*10**2,\"*10**-2 Nm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Maximum torque on the dipole is 1.2 *10**-2 Nm\n"
+ ]
+ }
+ ],
+ "prompt_number": 55
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.14 Page no 62"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "q=1*10**-6\n",
+ "a=2*10**-2\n",
+ "E=10**5\n",
+ "\n",
+ "#Calculation\n",
+ "p=q*a\n",
+ "W=2*p*E\n",
+ "\n",
+ "#Result\n",
+ "print\"Work done in the rotation is \", W*10**3,\"*10**-3 J\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Work done in the rotation is 4.0 *10**-3 J\n"
+ ]
+ }
+ ],
+ "prompt_number": 60
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.15 Page no 62"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "q=2*10**-6\n",
+ "a=0.1\n",
+ "m=9*10**9\n",
+ "r=0.5\n",
+ "\n",
+ "#Calculation\n",
+ "p=q*a\n",
+ "E=m*p/r**3\n",
+ "\n",
+ "#Result\n",
+ "print\"Electric field intensity is \",E*10**-4,\"*10**4 N/C\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Electric field intensity is 1.44 *10**4 N/C\n"
+ ]
+ }
+ ],
+ "prompt_number": 65
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.16 Page no 62"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "qa=2.5*10**-7\n",
+ "qb=-2.5*10**-7\n",
+ "a=15\n",
+ "b=15\n",
+ "\n",
+ "#Calculation\n",
+ "q=qa+qb\n",
+ "C=(a+b)*10**-2\n",
+ "E=qa*C\n",
+ "\n",
+ "#Result\n",
+ "print\"Total charge is \", q,\"\\nElectric dipole moment of the system is \",E,\"Cm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Total charge is 0.0 \n",
+ "Electric dipole moment of the system is 7.5e-08 Cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 71
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.17 Page no 63"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "p=2*10**-8\n",
+ "m=9*10**9\n",
+ "r=1\n",
+ "\n",
+ "#Calculation\n",
+ "E=(m*p*math.sqrt(3*(math.cos**2(60)*180/3.14))+1)/r**3\n",
+ "print E"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "ename": "TypeError",
+ "evalue": "'int' object is not callable",
+ "output_type": "pyerr",
+ "traceback": [
+ "\u001b[1;31m---------------------------------------------------------------------------\u001b[0m\n\u001b[1;31mTypeError\u001b[0m Traceback (most recent call last)",
+ "\u001b[1;32m<ipython-input-77-6c8b884fd561>\u001b[0m in \u001b[0;36m<module>\u001b[1;34m()\u001b[0m\n\u001b[0;32m 5\u001b[0m \u001b[1;33m\u001b[0m\u001b[0m\n\u001b[0;32m 6\u001b[0m \u001b[1;31m#Calculation\u001b[0m\u001b[1;33m\u001b[0m\u001b[1;33m\u001b[0m\u001b[0m\n\u001b[1;32m----> 7\u001b[1;33m \u001b[0mE\u001b[0m\u001b[1;33m=\u001b[0m\u001b[1;33m(\u001b[0m\u001b[0mm\u001b[0m\u001b[1;33m*\u001b[0m\u001b[0mp\u001b[0m\u001b[1;33m*\u001b[0m\u001b[0mmath\u001b[0m\u001b[1;33m.\u001b[0m\u001b[0msqrt\u001b[0m\u001b[1;33m(\u001b[0m\u001b[1;36m3\u001b[0m\u001b[1;33m*\u001b[0m\u001b[1;33m(\u001b[0m\u001b[0mmath\u001b[0m\u001b[1;33m.\u001b[0m\u001b[0mcos\u001b[0m\u001b[1;33m**\u001b[0m\u001b[1;36m2\u001b[0m\u001b[1;33m(\u001b[0m\u001b[1;36m60\u001b[0m\u001b[1;33m)\u001b[0m\u001b[1;33m*\u001b[0m\u001b[1;36m180\u001b[0m\u001b[1;33m/\u001b[0m\u001b[1;36m3.14\u001b[0m\u001b[1;33m)\u001b[0m\u001b[1;33m)\u001b[0m\u001b[1;33m+\u001b[0m\u001b[1;36m1\u001b[0m\u001b[1;33m)\u001b[0m\u001b[1;33m/\u001b[0m\u001b[0mr\u001b[0m\u001b[1;33m**\u001b[0m\u001b[1;36m3\u001b[0m\u001b[1;33m\u001b[0m\u001b[0m\n\u001b[0m\u001b[0;32m 8\u001b[0m \u001b[1;32mprint\u001b[0m \u001b[0mE\u001b[0m\u001b[1;33m\u001b[0m\u001b[0m\n",
+ "\u001b[1;31mTypeError\u001b[0m: 'int' object is not callable"
+ ]
+ }
+ ],
+ "prompt_number": 77
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.18 Page no 63"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "p=5*10**-8\n",
+ "m=9*10**9\n",
+ "r=0.15\n",
+ "\n",
+ "#Calculation\n",
+ "E=m*2*p/r**3\n",
+ "E1=m*p/r**3\n",
+ "\n",
+ "print\"(i) Electric field along AB is \", round(E*10**-5,2),\"*10**5 N/C\"\n",
+ "print\"(ii) Electric field along BA is \", round(E1*10**-5,2),\"*10**5 N/C\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Electric field along AB is 2.67 *10**5 N/C\n",
+ "(ii) Electric field along BA is 1.33 *10**5 N/C\n"
+ ]
+ }
+ ],
+ "prompt_number": 89
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter3_1.ipynb b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter3_1.ipynb new file mode 100644 index 00000000..8aff7044 --- /dev/null +++ b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter3_1.ipynb @@ -0,0 +1,1016 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:b7451f8a5d88007ceae262e7f31a9c2ef46acd4d68ca53e0c7b4ce898e623cd8"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 3 Electrostatic potential and flux"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.1 Page no 89"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "q=300*10**-6 #c\n",
+ "V=6\n",
+ "\n",
+ "#Calculation\n",
+ "W=q*V\n",
+ "\n",
+ "#Result\n",
+ "print\"Work done is \", W*10**3,\"*10**-3 J\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Work done is 1.8 *10**-3 J\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.2 Page no 89"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#given \n",
+ "Va=-10 #V\n",
+ "W=300 #J\n",
+ "q=3.0 #C\n",
+ "\n",
+ "#Calculation\n",
+ "V=(W/q)+Va\n",
+ "\n",
+ "#Result\n",
+ "print\"The value of V is \", V,\"Volts\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of V is 90.0 Volts\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.3 Page no 89"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "m=9*10**9\n",
+ "q=16*10**-10 #C\n",
+ "r=0.1\n",
+ "r1=0.06\n",
+ "q1=12*10**-10\n",
+ "\n",
+ "#Calculation\n",
+ "Vb=m*q/r\n",
+ "Vb1=m*q/r1\n",
+ "V=Vb1-Vb\n",
+ "W=q1*V\n",
+ "\n",
+ "#Result\n",
+ "print\"Workdone is \", W*10**8,\"*10**-8 J\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Workdone is 11.52 *10**-8 J\n"
+ ]
+ }
+ ],
+ "prompt_number": 17
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.4 Page no 89"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "r=3.4*10**-14 #m\n",
+ "n=47\n",
+ "q=1.6*10**-19 #C\n",
+ "m=9*10**9\n",
+ "\n",
+ "#Calculation\n",
+ "V=m*n*q/r\n",
+ "\n",
+ "#Result\n",
+ "print\"Electric potential at the surface of silver nucleus is \", round(V*10**-6,2),\"*10**6 V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Electric potential at the surface of silver nucleus is 1.99 *10**6 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 22
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.5 Page no 90"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "m=9*10**9\n",
+ "q=4*10**-6\n",
+ "\n",
+ "#Calculation\n",
+ "V=2*q*m\n",
+ "\n",
+ "#Result\n",
+ "print\"Electric potential is \", V*10**-3,\"*10**3 V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Electric potential is 72.0 *10**3 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 26
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.9 Page no 91"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "m=9*10**9\n",
+ "q=250*10**-6\n",
+ "r=0.1\n",
+ "\n",
+ "#Calculation\n",
+ "V=m*q/r\n",
+ "\n",
+ "#Result\n",
+ "print\"Electric potential at the centre is \", V*10**-7,\"*10**7 V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Electric potential at the centre is 2.25 *10**7 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 30
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.10 Page no 96"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "m=3*10**-16\n",
+ "g=9.8\n",
+ "d=5*10**-3\n",
+ "q=16.0*10**-18\n",
+ "\n",
+ "#Calculation\n",
+ "V=(m*g*d/q)*10\n",
+ "\n",
+ "#Result\n",
+ "print\"Voltage needed to balance an oil drop is \",round(V,2),\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Voltage needed to balance an oil drop is 9.19 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 38
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.12 Page no 96"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "q=1.6*10**-19 #C\n",
+ "V=3000 #V\n",
+ "r=5*10**-2 #m\n",
+ "g=9.8\n",
+ "\n",
+ "#Calculation\n",
+ "E=V/r\n",
+ "m=q*E/g\n",
+ "\n",
+ "#Result\n",
+ "print\"The mass of the particle is \", round(m*10**16,1),\"*10**-16 Kg\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The mass of the particle is 9.8 *10**-16 Kg\n"
+ ]
+ }
+ ],
+ "prompt_number": 44
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.13 Page no 100"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "m=9*10**-9\n",
+ "q1=3*10**-9\n",
+ "q2=3*10**-9\n",
+ "q3=10**9\n",
+ "r=0.2\n",
+ "\n",
+ "#Calculation\n",
+ "W=m*((q1*q3/r)+(q2*q3/r))\n",
+ "\n",
+ "#Result\n",
+ "print\"Workdone is \", W,\"J\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Workdone is 2.7e-07 J\n"
+ ]
+ }
+ ],
+ "prompt_number": 48
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.14 Page no 100"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "m=9*10**9\n",
+ "q=1.6*10**-19\n",
+ "r=10**-10\n",
+ "\n",
+ "#Calculation\n",
+ "U=m*q**2/r\n",
+ "K=U/2.0\n",
+ "\n",
+ "#Result\n",
+ "print\"Kinetic energy is \",K,\"J\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Kinetic energy is 1.152e-18 J\n"
+ ]
+ }
+ ],
+ "prompt_number": 53
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.15 Page no 100"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "m=9*10**-31\n",
+ "V=10**6\n",
+ "q=1.6*10**-19\n",
+ "a=9*10**9\n",
+ "\n",
+ "#Calculation\n",
+ "K=m*V**2\n",
+ "r=a*q**2/K\n",
+ "\n",
+ "#Result\n",
+ "print\"Distance of the closest approach is \", r,\"m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Distance of the closest approach is 2.56e-10 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 57
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.17 Page no 101"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "r=0.53*10**-10 #m\n",
+ "q1=1.6*10**-19 #C\n",
+ "q2=-1.6*10**-19 #C\n",
+ "a=9*10**9\n",
+ "r1=1.06*10**-10\n",
+ "\n",
+ "#Calculation\n",
+ "U=a*q1*q2/r\n",
+ "Ue=U/q1\n",
+ "K=-Ue/2.0\n",
+ "E=Ue+K\n",
+ "U1=(a*q1*q2/r1)/q1\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Potential energy of the system is \", round(Ue,1),\"eV\"\n",
+ "print\"(ii) Minimum amount of work required to free the elctrons ia \",round(E,1),\"ev\"\n",
+ "print\"(iii) Potential energyof the system is \",round(E,1) ,\"ev and work requiredto free the electrons is \",round(-E,1),\"eV\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Potential energy of the system is -27.2 eV\n",
+ "(ii) Minimum amount of work required to free the elctrons ia -13.6 ev\n",
+ "(iii) Potential energyof the system is -13.6 ev and work requiredto free the electrons is 13.6 eV\n"
+ ]
+ }
+ ],
+ "prompt_number": 75
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.18 Page no 102"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=9*10**9\n",
+ "q1=7*10**-6 #C\n",
+ "q2=-2*10**-6\n",
+ "r=0.18\n",
+ "r1=0.09\n",
+ "A=9*10**5\n",
+ "\n",
+ "#Calculation\n",
+ "U=a*q1*q2/r\n",
+ "W=0-U\n",
+ "U1=(q1*A/r1)+(q2*A/r1)+U\n",
+ "\n",
+ "#Result\n",
+ "print\"(a) Electrostatic potential energy is \", round(U,1),\"J\"\n",
+ "print\"(b) Work required to seperate two charges is \",round(W,1),\"J\"\n",
+ "print\"(c) Electrostatic energy is \", U1,\"J\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) Electrostatic potential energy is -0.7 J\n",
+ "(b) Work required to seperate two charges is 0.7 J\n",
+ "(c) Electrostatic energy is 49.3 J\n"
+ ]
+ }
+ ],
+ "prompt_number": 93
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.20 Page no 103"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "p=6*10**-6\n",
+ "E=10**6\n",
+ "a=1\n",
+ "\n",
+ "#Calculation,\n",
+ "U1=-p*E*a\n",
+ "U2=(p*E*(math.cos(60)*180/3.14))*10**-2\n",
+ "U3=U2-U1\n",
+ "\n",
+ "#Result\n",
+ "print\"Heat released by substance is \", round(U3,0),\"J\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Heat released by substance is 3.0 J\n"
+ ]
+ }
+ ],
+ "prompt_number": 103
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.21 Page no 109"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "q=10**-7\n",
+ "e=8.854*10**-12\n",
+ "\n",
+ "#Calculation\n",
+ "a=q/e\n",
+ "\n",
+ "#Result\n",
+ "print\"Electric flux through the surface of the cube is \", round(a*10**-4,2),\"Nm**2C-1\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Electric flux through the surface of the cube is 1.13 Nm**2C-1\n"
+ ]
+ }
+ ],
+ "prompt_number": 108
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.22 Page no 109"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "q=8.85*10**-6 \n",
+ "e=8.85*10**-12\n",
+ "\n",
+ "#Calculation\n",
+ "a=q/e\n",
+ "b=a/6.0\n",
+ "\n",
+ "#Result\n",
+ "print\"Electric flux through each face is \", round(b*10**-5,2),\"Nm**2C-1\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Electric flux through each face is 1.67 Nm**2C-1\n"
+ ]
+ }
+ ],
+ "prompt_number": 115
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.23 Page no 109"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "E0=2*10**3 #N/C\n",
+ "S=0.2\n",
+ "\n",
+ "#Calculation\n",
+ "a=(3/5.0)*E0*S\n",
+ "\n",
+ "#Result\n",
+ "print\"Electric flux of the field is \", a,\"Nm**2C-1\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Electric flux of the field is 240.0 Nm**2C-1\n"
+ ]
+ }
+ ],
+ "prompt_number": 118
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.24 Page no 109"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "r=0.2\n",
+ "m=9*10**9\n",
+ "b=50\n",
+ "\n",
+ "import math\n",
+ "E=250*r\n",
+ "a=E*4*math.pi*r**2\n",
+ "q=b*r**2/m\n",
+ "\n",
+ "#Result\n",
+ "print\"Charge contained in a sphere is \", round(q*10**10,2)*10**-10,\"C\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Charge contained in a sphere is 2.22e-10 C\n"
+ ]
+ }
+ ],
+ "prompt_number": 128
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.25 Page no 110"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=0.1 #m\n",
+ "A=800\n",
+ "e=8.854*10**-12\n",
+ "\n",
+ "#Calculation\n",
+ "b=A*a**2.5*(math.sqrt(2)-1)\n",
+ "q=e*b\n",
+ "\n",
+ "#Result\n",
+ "print\"(a) The flux through the cube is \", round(b,2),\"Nm**2C-1\"\n",
+ "print\"The charge within the cube is \",round(q*10**12,2)*10**-12,\"C\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) The flux through the cube is 1.05 Nm**2C-1\n",
+ "The charge within the cube is 9.28e-12 C\n"
+ ]
+ }
+ ],
+ "prompt_number": 139
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.26 Page no 111"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "E=200\n",
+ "a=0.05\n",
+ "e=8.854*10**-12\n",
+ "d=3.14\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "b=E*math.pi*a**2\n",
+ "c=2*b\n",
+ "q=e*d\n",
+ "\n",
+ "#Result\n",
+ "print\"(a) Net outward flux through each flat face is \", round(b,2),\"Nm**2C-1\"\n",
+ "print\"(b) Flux through the side of cylinder is zero \"\n",
+ "print\"(c) Net outward flux through the cylinder is \", round(c,2),\"Nm**2C-1\"\n",
+ "print\"(d) The net charge in the cylinder is \",round(q*10**11,2)*10**-11,\"C\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) Net outward flux through each flat face is 1.57 Nm**2C-1\n",
+ "(b) Flux through the side of cylinder is zero \n",
+ "(c) Net outward flux through the cylinder is 3.14 Nm**2C-1\n",
+ "(d) The net charge in the cylinder is 2.78e-11 C\n"
+ ]
+ }
+ ],
+ "prompt_number": 156
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.28 Page no 114"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "q=5.8*10**-6 #C\n",
+ "r=8*10**-2 #m\n",
+ "e=8.854*10**-12\n",
+ "l=3.0\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "E=q/(2*math.pi*e*r*l)\n",
+ "\n",
+ "#Result\n",
+ "print\"Electric field is \", round(E*10**-5,1),\"*10**5 N/C\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Electric field is 4.3 *10**5 N/C\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.29 Page no 114"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "E=9*10**4 #N/C\n",
+ "r=2*10**-2 #m\n",
+ "m=9*10**9\n",
+ "\n",
+ "#Calculation\n",
+ "a=r*E/(2.0*m)\n",
+ "print\"Linear charge density is \", a,\"Cm-1\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Linear charge density is 1e-07 Cm-1\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.30 Page no 115"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "q=10*10**-6 #C\n",
+ "r=0.1 #m\n",
+ "a=8.85*10**-12\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "E=q/(4.0*math.pi*a*r**2)\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Electric field intensity at a point 10 cm from the centre\", round(E*10**-6,0),\"*10**6 N/C\"\n",
+ "print\"(ii) Since the point is lying inside the shell, electric intensity at this point is zero\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Electric field intensity at a point 10 cm from the centre 9.0 *10**6 N/C\n",
+ "(ii) Since the point is lying inside the shell, electric intensity at this point is zero\n"
+ ]
+ }
+ ],
+ "prompt_number": 17
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.31 Page no 118"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Z=79\n",
+ "e=1.6*10**-19\n",
+ "e0=8.854*10**-12\n",
+ "R=6.2*10**-15\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "q=Z*e\n",
+ "E=q/(4.0*math.pi*e0*R**2)\n",
+ "b=E/4.0\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) The magnitude of the electric field at the surface of nucleus is \", round(E*10**-21,0)*10**21,\"N/C\"\n",
+ "print\"(ii) The magnitude of the electric field at a distance 2R from the centre of the nucleus is \",round(b*10**-21,2),\"*10**21 N/C\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) The magnitude of the electric field at the surface of nucleus is 3e+21 N/C\n",
+ "(ii) The magnitude of the electric field at a distance 2R from the centre of the nucleus is 0.74 *10**21 N/C\n"
+ ]
+ }
+ ],
+ "prompt_number": 31
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.32 Page no 119"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "e=8.854*10**-12\n",
+ "A=0.5\n",
+ "F=1.8*10**-12 #N\n",
+ "E=1.6*10**-19\n",
+ "\n",
+ "#Calculation\n",
+ "q=(2*e*A**2*F)/E\n",
+ "\n",
+ "#Result\n",
+ "print\"Total charge on the sheet is \", round(q*10**6,0),\"micro C\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Total charge on the sheet is 50.0 micro C\n"
+ ]
+ }
+ ],
+ "prompt_number": 43
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.33 Page no 119"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=5*10**-6\n",
+ "e=8.854*10**-12\n",
+ "r=0.1\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "b=-(a*math.pi*r**2*(math.cos(60)*180/3.14))/(2*e)\n",
+ "\n",
+ "#Result\n",
+ "print\"Electric flux through a circular area is \", round(b*10**-5,2),\"*10**3 Nm**2C-1\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Electric flux through a circular area is 4.84 *10**3 Nm**2C-1\n"
+ ]
+ }
+ ],
+ "prompt_number": 54
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter4_1.ipynb b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter4_1.ipynb new file mode 100644 index 00000000..5987ff7f --- /dev/null +++ b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter4_1.ipynb @@ -0,0 +1,1326 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:2ebe494197bc592ac147978ecceacaa802bf7a7b9283aeec109e01967ce4cfa8"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 4 Capacitance "
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.1 Page no 159"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "m=9*10**9\n",
+ "r=6.4*10**6 #m\n",
+ "\n",
+ "#Calculation\n",
+ "C=r/m\n",
+ "\n",
+ "#Result\n",
+ "print\"The capacitance of the earth is \", round(C*10**6,0),\"micro F\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The capacitance of the earth is 711.0 micro F\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.2 Page no 160"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "m=9*10**9\n",
+ "c=50*10**-12\n",
+ "V=10**4\n",
+ "\n",
+ "#Calculation\n",
+ "r=(m*c)*10**2\n",
+ "q=(c*V)\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Radius of a isolated sphere is \",r,\"cm\"\n",
+ "print\"(ii) Charge of a isolated sphere is \", q*10**6,\"micro C\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Radius of a isolated sphere is 45.0 cm\n",
+ "(ii) Charge of a isolated sphere is 0.5 micro C\n"
+ ]
+ }
+ ],
+ "prompt_number": 21
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.3 Page no 160"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "r=3*10**-3 #m\n",
+ "m=9*10**9\n",
+ "q1=27*10**-12 #C\n",
+ "\n",
+ "#Calculation\n",
+ "R=3*r\n",
+ "C=R/m\n",
+ "V=q1/C\n",
+ "\n",
+ "#Result\n",
+ "print\"Capacitance of the bigger drop is \", C*10**12,\"pico F \\npotential of the bigger drop is \",V,\"Volts\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Capacitance of the bigger drop is 1.0 pico F \n",
+ "potential of the bigger drop is 27.0 Volts\n"
+ ]
+ }
+ ],
+ "prompt_number": 33
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.4 Page no 162"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "m=9*10**9\n",
+ "ra=0.09\n",
+ "rb=0.1\n",
+ "\n",
+ "#Calculation\n",
+ "C=ra*rb/(m*(rb-ra))\n",
+ "\n",
+ "#Result\n",
+ "print\"Capacitance of the capacitor is \", C*10**12,\"pico F\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Capacitance of the capacitor is 100.0 pico F\n"
+ ]
+ }
+ ],
+ "prompt_number": 37
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.5 Page no 162"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "r=2 #cm\n",
+ "d=1.2\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "R=(d/r)*10**2\n",
+ "rab=(R*2)\n",
+ "x=r**2+4*rab\n",
+ "y=math.sqrt(x)\n",
+ "\n",
+ "#Result\n",
+ "print\"ra+rb=\", y,\"cm \\nra-rb=\",r ,\"cm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "ra+rb= 22.0 cm \n",
+ "ra-rb= 2 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 64
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.6 Page no 164"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "d=10**-3 #m\n",
+ "c=1 #F\n",
+ "e=8.854*10**-12\n",
+ "\n",
+ "#Calculation\n",
+ "A=c*d/e\n",
+ "\n",
+ "#Result\n",
+ "print\"Area is \", round(A*10**-8,1),\"*10**8 m**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Area is 1.1 *10**8 m**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 69
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.7 Page no 164"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "A=0.02 #m**2\n",
+ "r=0.5 #m\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "d=A/(4.0*math.pi*r)\n",
+ "\n",
+ "#Result\n",
+ "print\"Distance is \", round(d*10**3,2),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Distance is 3.18 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 74
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.8 Page no 164"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "e=8.854*10**-12\n",
+ "K=6\n",
+ "A=30\n",
+ "d=2.0*10**-3\n",
+ "E=500\n",
+ "\n",
+ "#Calculation\n",
+ "C=e*K*A/d\n",
+ "V=E*d*10**3\n",
+ "q=C*V\n",
+ "\n",
+ "#Result\n",
+ "print\"Capacitance of a parallel plate \", round(q*10**3,3),\"micro C\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Capacitance of a parallel plate 0.797 micro C\n"
+ ]
+ }
+ ],
+ "prompt_number": 87
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.9 Page no 165"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "C=300*10**-12\n",
+ "V=10*10**3\n",
+ "A=0.01\n",
+ "d=1*10**-3\n",
+ "\n",
+ "#Calculation\n",
+ "q=C*V\n",
+ "a=q/A\n",
+ "E=V/d\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Charge on each plate is \", q,\"C\"\n",
+ "print\"(ii) Electric flux density is \", a*10**4,\"10**-4 C/m**2\"\n",
+ "print\"(iii) Potential gradient is \", E,\"V/m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Charge on each plate is 3e-06 C\n",
+ "(ii) Electric flux density is 3.0 10**-4 C/m**2\n",
+ "(iii) Potential gradient is 10000000.0 V/m\n"
+ ]
+ }
+ ],
+ "prompt_number": 101
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.10 Page no 165"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "A2=500 #cm**2\n",
+ "A1=100 #cm**2\n",
+ "d1=0.05 #cm\n",
+ "\n",
+ "#Calculation\n",
+ "d2=A2*d1/A1\n",
+ "\n",
+ "#Result\n",
+ "print\"Distance between the plates of second capacitor is \", d2,\"cm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Distance between the plates of second capacitor is 0.25 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 104
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.11 page no 167"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "c1=0.5 #micro F\n",
+ "c2=0.3 #micro F\n",
+ "c3=0.2 #micro F\n",
+ "\n",
+ "#Calculation\n",
+ "Cp=c1+c2+c3 \n",
+ "Cs=(1/c1)+(1/c2)+(1/c3)\n",
+ "\n",
+ "#Result\n",
+ "print\" The ratio ofmaximum capacitance to minimum capacitance is \",round (Cs,1)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " The ratio ofmaximum capacitance to minimum capacitance is 10.3\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.12 Page no 168"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "c1=15.0 #micro F\n",
+ "c2=20.0 #micro F\n",
+ "V=10**-6\n",
+ "v1=600 #V \n",
+ "\n",
+ "#Calculation\n",
+ "Cs=c1*c2/(c1+c2)\n",
+ "Q=Cs*V*v1\n",
+ "Pd=(Q/c1)*10**6\n",
+ "Pd1=(Q/c2)*10**6\n",
+ "\n",
+ "#Result\n",
+ "print\"(i)charge on each capacitor is\",round(Q *10**3,2),\"10**-3 C\"\n",
+ "print\"(ii)P.D across15 micro Fcapacitor is\",round (Pd,1),\"V\"\n",
+ "print\" P.D across 20 micro F is\",round (Pd1,0),\"V\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i)charge on each capacitor is 5.14 10**-3 C\n",
+ "(ii)p.D across15 micro Fcapacitor is 342.9 V\n",
+ " P.D across 20 micro F is 257.0 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 30
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.13 page no.168"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Ca=18 #micro F\n",
+ "Cb=4 #micro F\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "C=Ca*Cb\n",
+ "C12=math.sqrt(Ca**2-4*C)\n",
+ "C2=2*C12\n",
+ "\n",
+ "#Result\n",
+ "print\"The capacitance of capacitor C1 is\", C12,\"micro F\"\n",
+ "print\"The capacitance of capacitor C2 is\",C2,\"micro F\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The capacitance of capacitor C1 is 6.0 micro F\n",
+ "The capacitance of capacitor C2 is 12.0 micro F\n"
+ ]
+ }
+ ],
+ "prompt_number": 39
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.14 Page no 168"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "q=750*10**-6\n",
+ "C1=15*10**-6\n",
+ "V2=20.0 #V\n",
+ "C3=8*10**-6\n",
+ "\n",
+ "#Calculation\n",
+ "V1=q/C1\n",
+ "V=V1+V2\n",
+ "q3=C3*V2\n",
+ "q2=q-q3\n",
+ "C2=q2/V2\n",
+ "\n",
+ "#Result\n",
+ "print\"The value of V1 is \", V1,\"V\"\n",
+ "print\"The value of V is \",V,\"V\"\n",
+ "print\"The value of capacitance is\",C2*10**6,\"micro F\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of V1 is 50.0 V\n",
+ "The value of V is 70.0 V\n",
+ "The value of capacitance is 29.5 micro F\n"
+ ]
+ }
+ ],
+ "prompt_number": 52
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.15 Page no 169"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "C2=9.0 #micro F\n",
+ "C3=9.0\n",
+ "C4=9.0\n",
+ "C1=3\n",
+ "V=10 #V\n",
+ "\n",
+ "#Calculation\n",
+ "C=1/((1/C2)+(1/C3)+(1/C4))\n",
+ "Cab=C1+C\n",
+ "q=Cab*V\n",
+ "\n",
+ "#Result\n",
+ "print\"Equivalent capacitance between point A and B is \", Cab,\"micro F\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Equivalent capacitance between point A and B is 6.0 micro F\n"
+ ]
+ }
+ ],
+ "prompt_number": 60
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.17 Page no 169"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Cab=10 #micro F\n",
+ "C1=8.0 #micro F\n",
+ "C2=8.0\n",
+ "C3=8\n",
+ "C4=8\n",
+ "C5=12\n",
+ "V=400\n",
+ "\n",
+ "#Calculation\n",
+ "Cbc=((C1*C2)/(C1+C2))+C3+C4\n",
+ "Cac=Cab*Cbc/(Cab+Cbc)\n",
+ "Ccd=C1+C5\n",
+ "Cad=Cac*Ccd/(Cac+Ccd)\n",
+ "q=Cad*V\n",
+ "Vcd=q/Ccd\n",
+ "q1=C5*Vcd\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) The equivalent capacitance between A and D is \", Cad,\"micro f\"\n",
+ "print\"(ii) The charge on 12 micro F capacitor is \",q1*10**-3,\"mC\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) The equivalent capacitance between A and D is 5.0 micro f\n",
+ "(ii) The charge on 12 micro F capacitor is 1.2 mC\n"
+ ]
+ }
+ ],
+ "prompt_number": 84
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.20 Page no 171"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "C1=5 #micro F\n",
+ "C2=6 #micro F\n",
+ "V=10 #V\n",
+ "\n",
+ "#Calculation\n",
+ "Cp=C1+C2\n",
+ "q=Cp*V\n",
+ "\n",
+ "#Result\n",
+ "print\"Charge supplied by battery is \", q,\"micro F\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Charge supplied by battery is 110 micro F\n"
+ ]
+ }
+ ],
+ "prompt_number": 88
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.21 Page no 171"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "C1=2 #micro F\n",
+ "C2=2 #micro F\n",
+ "C3=2\n",
+ "C4=2\n",
+ "\n",
+ "#Calculation\n",
+ "Cs=C1*C2/(C1+C2)\n",
+ "Cab=C3*C4/(C3+C4)\n",
+ "\n",
+ "#Result\n",
+ "print\"The capacitance of the Capacitors\", Cab,\"micro F\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The capacitance of the Capacitors 1 micro F\n"
+ ]
+ }
+ ],
+ "prompt_number": 93
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.22 Page no 171"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "C1=10.0 #micro F\n",
+ "C2=10.0\n",
+ "C3=10.0\n",
+ "C4=10*10**-3\n",
+ "V=500 #V\n",
+ "\n",
+ "#Calculation\n",
+ "Cs=1/((1/C1)+(1/C2)+(1/C3))\n",
+ "Cab=Cs+(C4*10**3)\n",
+ "Q=(C1*(500/3.0))*10**-3\n",
+ "Q1=C4*V\n",
+ "\n",
+ "#Result\n",
+ "print\"(a) The equivalent capacitance of the network is\",round(Cab,1),\"micro F\"\n",
+ "print \"(b) The charge on 12 micro F Capacitor is\",Q1,\"*10**-3 C\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) The equivalent capacitance of the network is 13.3 micro F\n",
+ "(b) The charge on 12 micro F Capacitor is 5.0 *10**-3 C\n"
+ ]
+ }
+ ],
+ "prompt_number": 115
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.23 Page no 172"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "C4=6 #micro F\n",
+ "C5=12 \n",
+ "C1=8.0\n",
+ "C7=1\n",
+ "\n",
+ "#Calculation\n",
+ "Cs=C4*C5/(C4+C5)\n",
+ "C11=(C1*Cs)/(C1+Cs)\n",
+ "Cs1=C1*C7/(C1+C7)\n",
+ "Cp=C11+Cs1\n",
+ "C=1/(1-(1/Cp))\n",
+ "\n",
+ "#Result\n",
+ "print\"The value of capacitance C is \", round(C,2),\"micro F\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of capacitance C is 1.39 micro F\n"
+ ]
+ }
+ ],
+ "prompt_number": 129
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.24 Page no 175"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "K=5\n",
+ "l=0.2\n",
+ "c=10**-9 #F\n",
+ "b=15.4\n",
+ "a=15\n",
+ "pd=5000 #V\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "C=(K*l*c)/(41.1*math.log10(b/a))\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) The capacitance of cylindrical capacitor is \", round(C*10**9,1)*10**-9,\"F\"\n",
+ "print\"(ii) The potential of the inner cylinder is equal to p.d. between two cylinders i.e potentila of inner cylinder is\",pd,\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) The capacitance of cylindrical capacitor is 2.1e-09 F\n",
+ "(ii) The potential of the inner cylinder is equal to p.d. between two cylinders i.e potentila of inner cylinder is 5000 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 138
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.25 Page no 179"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "C=5*10**-6\n",
+ "V=100\n",
+ "C1=3*10**-6\n",
+ "\n",
+ "#Calculation\n",
+ "q=C*V\n",
+ "Cp=C+C1\n",
+ "pd=q/Cp\n",
+ "\n",
+ "#Result\n",
+ "print\"P.D across the capacitor is \", pd,\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "P.D across the capacitor is 62.5 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 143
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.26 Page no 179 "
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "V=250 #V\n",
+ "C1=6 #micro F\n",
+ "C2=4\n",
+ "Cp=10*10**-6\n",
+ "\n",
+ "#Calculation\n",
+ "pd=V*C1/(C1+C2)\n",
+ "q=pd*C2*10**-6\n",
+ "q1=2*q\n",
+ "pd1=q1/Cp\n",
+ "q2=C2*pd1\n",
+ "q3=C1*pd1\n",
+ "\n",
+ "#Result\n",
+ "print\"New potentila difference is \", pd1,\"V\"\n",
+ "print\"Charge on 4 micro F capacitor is \",q2,\"micro C\"\n",
+ "print\"Charge on 6 micro F capacitor is \",q3,\"micro C\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "New potentila difference is 120.0 V\n",
+ "Charge on 4 micro F capacitor is 480.0 micro C\n",
+ "Charge on 6 micro F capacitor is 720.0 micro C\n"
+ ]
+ }
+ ],
+ "prompt_number": 156
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.28 Page no 180"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "C1=16*10**-6 # F\n",
+ "C2=4 #micro F\n",
+ "V1=100 #V\n",
+ "Cp=20*10**-6 #f\n",
+ "\n",
+ "#Calculation\n",
+ "q=C1*V1\n",
+ "U1=0.5*C1*V1**2\n",
+ "V=q/Cp\n",
+ "U2=0.5*Cp*V**2\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Potential difference across the capacitor is \", V,\"Volts\"\n",
+ "print\"(ii) The electrostatic energies before and after the capacitors are connected \",U2,\"J\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Potential difference across the capacitor is 80.0 Volts\n",
+ "(ii) The electrostatic energies before and after the capacitors are connected 0.064 J\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.29 Page no 180"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "m=9*10**9\n",
+ "V=3.0*10**6\n",
+ "r=2\n",
+ "\n",
+ "#Calculation\n",
+ "q=(V*r)/m\n",
+ "E=0.5*q*V\n",
+ "\n",
+ "#Result\n",
+ "print\"The heat generated is \", E,\"J\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The heat generated is 1000.0 J\n"
+ ]
+ }
+ ],
+ "prompt_number": 19
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.30 Page no 180"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "V=12 #V\n",
+ "C=1.35*10**-10 #C\n",
+ "\n",
+ "#Calculation\n",
+ "q=C\n",
+ "\n",
+ "#Result\n",
+ "print\"Extra Charge supplied by battery is \", q,\"C\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Extra Charge supplied by battery is 1.35e-10 C\n"
+ ]
+ }
+ ],
+ "prompt_number": 26
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.31 Page no 181"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "C=100*10**-6 #F\n",
+ "V=500 #V\n",
+ "\n",
+ "#Calculation\n",
+ "q=V/2.0\n",
+ "E=0.5*(0.5*C*V**2)\n",
+ "\n",
+ "#Result\n",
+ "print\"Charge in the new stored energy is \", E,\"J\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Charge in the new stored energy is 6.25 J\n"
+ ]
+ }
+ ],
+ "prompt_number": 31
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.32 Page no 187"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "A=2*10**-3 #m**2\n",
+ "d=0.01 #m\n",
+ "t=6*10**-3 #m\n",
+ "K=3\n",
+ "a=8.854*10**-12\n",
+ "\n",
+ "#Calculation\n",
+ "C=a*A/(d-t*(1-(1/3.0)))\n",
+ "\n",
+ "#Result\n",
+ "print\"The capacitance of the capacitor is \", round(C*10**12,2)*10**-12,\"F\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The capacitance of the capacitor is 2.95e-12 F\n"
+ ]
+ }
+ ],
+ "prompt_number": 37
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.33 Page no 188"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "e=8.854*10**-12\n",
+ "A=2\n",
+ "t1=0.5*10**-3\n",
+ "t2=1.5*10**-3\n",
+ "t3=0.3*10**-3\n",
+ "K1=2.0\n",
+ "K2=4.0\n",
+ "K3=6.0\n",
+ "\n",
+ "#Calculation\n",
+ "C=(e*A)/((t1/K1)+(t2/K2)+(t3/K3))\n",
+ "\n",
+ "#Result\n",
+ "print\"The capacitance of the capacitor is \", round(C*10**6,3)*10**-6,\"F\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The capacitance of the capacitor is 2.6e-08 F\n"
+ ]
+ }
+ ],
+ "prompt_number": 52
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.34 Page no 188"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=3 #mm\n",
+ "b=4.0 #mm\n",
+ "K1=5\n",
+ "\n",
+ "#Calaculation\n",
+ "K2=1/((a**2/b)-a/b)*K1\n",
+ "\n",
+ "#Result\n",
+ "print\"The relative permittivity of the additional dielectric is \", round(K2,2)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The relative permittivity of the additional dielectric is 3.33\n"
+ ]
+ }
+ ],
+ "prompt_number": 56
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.35 Page no 188"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "d=5\n",
+ "t=2\n",
+ "K=3.0\n",
+ "\n",
+ "#Calculation\n",
+ "D=d+(t-t/K)\n",
+ "\n",
+ "#Result\n",
+ "print\"New seperaion between the plates are \", round(D,2),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "New seperaion between the plates are 6.33 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 62
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.36 Page no 189"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "d=4\n",
+ "t=2\n",
+ "K=4.0\n",
+ "C1=50*10**-12 #f\n",
+ "V0=200 #V\n",
+ "\n",
+ "#Calculation\n",
+ "C=(d-t+(t/K))/d\n",
+ "q=C1*V0\n",
+ "V=V0*C\n",
+ "U=0.5*q*V\n",
+ "E=0.5*q*(V0-V)\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Final charge on ach plate is \", q,\"C\"\n",
+ "print\"(ii) P.D batween the plates is \", V,\"volts\"\n",
+ "print\"(iii)Final energy in the capacitor is \", U,\"J\"\n",
+ "print\"(iv) Energy loss is \", E,\"J\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Final charge on ach plate is 1e-08 C\n",
+ "(ii) P.D batween the plates is 125.0 volts\n",
+ "(iii)Final energy in the capacitor is 6.25e-07 J\n",
+ "(iv) Energy loss is 3.75e-07 J\n"
+ ]
+ }
+ ],
+ "prompt_number": 78
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.39 Page no 193"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "V=25*10**5\n",
+ "E=5.0*10**7\n",
+ "\n",
+ "#Calculation\n",
+ "r=V/E\n",
+ "\n",
+ "#Result\n",
+ "print\"Minimum radius of the spherical shell is \", r*100,\"cm\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Minimum radius of the spherical shell is 5.0 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 82
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter5_1.ipynb b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter5_1.ipynb new file mode 100644 index 00000000..639e5486 --- /dev/null +++ b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter5_1.ipynb @@ -0,0 +1,1741 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:76fc8177d7b8f6f9c96106003a3d3f70d960561edd9596ed73b03964026f4fb8"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 5 Electric current and resistance "
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.1 Page no 257"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "n=10**17\n",
+ "e=1.6*10**-19 #C\n",
+ "t=1.0 #S\n",
+ "\n",
+ "#Calculation\n",
+ "I=n*e/t\n",
+ "\n",
+ "#Result\n",
+ "print\"The magnitude of current in the wire is \",I*10**2,\"10**-2 A and direction is from left to right\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The magnitude of current in the wire is 1.6 10**-2 A and direction is from left to right\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.2 Page no 257"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#given\n",
+ "I=0.5\n",
+ "T=1\n",
+ "e=1.6*10**-19\n",
+ "t=60 #minute\n",
+ "\n",
+ "#Calculation\n",
+ "n=I*T/e\n",
+ "q=I*t**2\n",
+ "n1=q/e\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) The number of electrons passing a cross section of the bulb is \",round(n*10**-18,1)*10**18,\"electrons/S\"\n",
+ "print\"(ii) The number of electrons is \",round(n1*10**-22,1)*10**22,\"electrons/hour\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) The number of electrons passing a cross section of the bulb is 3.1e+18 electrons/S\n",
+ "(ii) The number of electrons is 1.1e+22 electrons/hour\n"
+ ]
+ }
+ ],
+ "prompt_number": 21
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.3 Page no 257"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "e=1.6*10**-19 #C\n",
+ "f=6.8*10**15 #rev/sec\n",
+ "r=0.51*10**-10 #m\n",
+ "\n",
+ "#Calculation\n",
+ "I=e*f\n",
+ "\n",
+ "#Result\n",
+ "print\"The equivalent current is \", I*10**3,\"*10**-3 A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The equivalent current is 1.088 *10**-3 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 25
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.5 Page no 267"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "I=10 #A\n",
+ "A=1 #m*m**2\n",
+ "e=1.6*10**-19 #C\n",
+ "n=10**28 #m**-3\n",
+ "\n",
+ "#Calculation\n",
+ "Vd=I/(n*A*e)\n",
+ "\n",
+ "#Result\n",
+ "print\"Drift velocity of the conduction electrons are \", Vd,\"m/s\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Drift velocity of the conduction electrons are 6.25e-09 m/s\n"
+ ]
+ }
+ ],
+ "prompt_number": 28
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.6 Page no 267"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "I=10 #A\n",
+ "A=4*10**-6 #m**2\n",
+ "e=1.6*10**-19 #C\n",
+ "n=8*10**28 #m**-3\n",
+ "l=4\n",
+ "\n",
+ "#Calculation\n",
+ "Vd=I/(n*A*e)\n",
+ "t=l/Vd\n",
+ "\n",
+ "#Result\n",
+ "print\"Time required by an electron is \", t*10**-4,\"*10**4 S\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Time required by an electron is 2.048 *10**4 S\n"
+ ]
+ }
+ ],
+ "prompt_number": 35
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.7 page no 267"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=6.023*10**23\n",
+ "m=63.5*10**-3\n",
+ "d=9*10**3\n",
+ "A=10**-7 #m**2\n",
+ "e=1.6*10**-19 #C\n",
+ "I=1.5 #a\n",
+ "K=1.38*10**-23 #J/K\n",
+ "T=300 #K\n",
+ "Vd=1.1*10**-3\n",
+ "C=3*10**8\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "n=a*d/m\n",
+ "Vd=I/(n*A*e)\n",
+ "V=math.sqrt((3*K*T*a)/m)\n",
+ "V1=Vd/V\n",
+ "E=Vd/C\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Thermal speeds of copper atoms at ordinary temperatures are \", round(V1*10**6,2),\"*10**-6 m/s\"\n",
+ "print\"(ii) Speed of propagation of electric fild is \", round(E*10**12,1)*10**-12"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Thermal speeds of copper atoms at ordinary temperatures are 3.2 *10**-6 m/s\n",
+ "(ii) Speed of propagation of electric fild is 3.7e-12\n"
+ ]
+ }
+ ],
+ "prompt_number": 53
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.8 Page no 268"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "V=5\n",
+ "l=0.1\n",
+ "Vd=2.5*10**-4\n",
+ "\n",
+ "#Calculation\n",
+ "E=V/l\n",
+ "u=Vd/E\n",
+ "\n",
+ "#Result\n",
+ "print\"The electron mobility is \", u,\"m**2/V/C\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The electron mobility is 5e-06 m**2/V/C\n"
+ ]
+ }
+ ],
+ "prompt_number": 57
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.9 Page no 268"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "I=2.4\n",
+ "A=0.30*10**-6\n",
+ "m=9.1*10**-31\n",
+ "n=8.4*10**28\n",
+ "e=1.6*10**-19\n",
+ "E=7.5\n",
+ "\n",
+ "#Calculation\n",
+ "J=I/A\n",
+ "t=m*J/(n*e**2*E)\n",
+ "\n",
+ "#Result\n",
+ "print\"Average relaxation time is \", round(t*10**16,2)*10**-16,\"S\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Average relaxation time is 4.51e-16 S\n"
+ ]
+ }
+ ],
+ "prompt_number": 64
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.10 Page no 268"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "r=0.12*10**-2 #m\n",
+ "I=10\n",
+ "r1=0.08*10**-2 #m\n",
+ "I=10 #A\n",
+ "e=1.6*10**-19 #C\n",
+ "n=8.4*10**28\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "A=math.pi*(r**2)\n",
+ "J=I/A\n",
+ "A1=math.pi*r1**2\n",
+ "Vd=I/(e*n*A1)\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Current density in the alluminium wire is \",round(J*10**-6,1),\"*10**6 A/m**2\"\n",
+ "print\"(ii) Drift velocity of electrons in the copper wire is \",round(Vd*10**4,1),\"*10**-4 m/S\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Current density in the alluminium wire is 2.2 *10**6 A/m**2\n",
+ "(ii) Drift velocity of electrons in the copper wire is 3.7 *10**-4 m/S\n"
+ ]
+ }
+ ],
+ "prompt_number": 82
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.11 Page no 269"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "D=0.13*10**-2\n",
+ "R=3.4 #ohms\n",
+ "l=10.0\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "A=(math.pi/4.0)*D**2\n",
+ "a=R*A/l\n",
+ "b=1/a\n",
+ "\n",
+ "#Result\n",
+ "print\"Conductivity of a material is \",round(b*10**-6,1),\"*10**6 S/m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Conductivity of a material is 2.2 *10**6 S/m\n"
+ ]
+ }
+ ],
+ "prompt_number": 93
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.12 Page no 269"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "A1=25.0 #mm**2\n",
+ "l2=1 #m\n",
+ "R2=1/58.0\n",
+ "A2=1\n",
+ "l1=1000\n",
+ "\n",
+ "#Calculation\n",
+ "R=(l1/l2)*(A2/A1)\n",
+ "R1=R*R2\n",
+ "\n",
+ "#Result\n",
+ "print\"The value of resistance is \", round(R1,2),\"ohm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of resistance is 0.69 ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 98
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.13 Page no 269"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R1=4.5\n",
+ "A1=1\n",
+ "A2=2.0\n",
+ "l2=3\n",
+ "l1=1.0\n",
+ "\n",
+ "#Calculation\n",
+ "R=(l2/l1)*(A1/A2)\n",
+ "R2=R*R1\n",
+ "\n",
+ "#Result\n",
+ "print\"The resistance of another wire is \", R2,\"ohm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The resistance of another wire is 6.75 ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 103
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.14 Page no 269"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "r=1\n",
+ "r1=0.5\n",
+ "R1=0.15 #ohm\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "A1=(math.pi/4.0)*r**2\n",
+ "A2=(math.pi/4.0)*r1**2\n",
+ "l=A1/A2\n",
+ "R=l*l\n",
+ "R2=R*R1\n",
+ "\n",
+ "#Result\n",
+ "print\"New resistance of the wire is \", R2,\"ohm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "New resistance of the wire is 2.4 ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 115
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.15 Page no 270"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R=1.5 #ohm\n",
+ "e=1.6*10**-19 #C\n",
+ "t=1 #second\n",
+ "V=3 #V\n",
+ "\n",
+ "#Calculation\n",
+ "I=V/R\n",
+ "n=I*t/e\n",
+ "\n",
+ "#Result\n",
+ "print \"Number of electrons flowing through it in 1 S is \",n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Number of electrons flowing through it in 1 S is 1.25e+19\n"
+ ]
+ }
+ ],
+ "prompt_number": 118
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.16 Page no 270"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "ne=2.8*10**18\n",
+ "np=1.2*10**18\n",
+ "e=1.6*10**-19\n",
+ "t=1 #S\n",
+ "V=220\n",
+ "\n",
+ "#Calculation\n",
+ "q=(ne+np)*e\n",
+ "I=q/t\n",
+ "R=V/I\n",
+ "\n",
+ "#Result\n",
+ "print\"Effective resistance of the tube is \", round(R,0),\"ohm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Effective resistance of the tube is 344.0 ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 124
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.17 Page no 270"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "m=84 #g\n",
+ "d=10.5 #g/cm**3\n",
+ "a=1.6*10**-6\n",
+ "\n",
+ "#Calculation\n",
+ "V=m/d\n",
+ "s=V**(1/3.0)\n",
+ "R=a/2.0\n",
+ "\n",
+ "#Result\n",
+ "print\"Resistance between the opposite faces is \", R,\"ohm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Resistance between the opposite faces is 8e-07 ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 130
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.18 Page no 270"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "l=1.001\n",
+ "A=1.001\n",
+ "\n",
+ "#Calculation\n",
+ "R=l*A\n",
+ "R1=R-1\n",
+ "A=R1*100\n",
+ "\n",
+ "#Result\n",
+ "print\"Percentage change in its resistance is \", round(A,1),\"%\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Percentage change in its resistance is 0.2 %\n"
+ ]
+ }
+ ],
+ "prompt_number": 137
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.19 Page no 271"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "m=0.45 #Kg\n",
+ "R=0.0014 #ohm\n",
+ "a=1.78*10**-8 #ohm\n",
+ "d=8.93*10**3 #Kg/m**3\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "l=math.sqrt(R*m/(a*d))\n",
+ "r=math.sqrt(m/(math.pi*d*1.99))\n",
+ "\n",
+ "#Result\n",
+ "print\"The value of length is\",round(l,2),\"m\"\n",
+ "print\"The value of radius is \",round(r*10**3,2),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of length is 1.99 m\n",
+ "The value of radius is 2.84 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 148
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.20 Page no 271"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R15=80 #ohm\n",
+ "a=0.004\n",
+ "\n",
+ "#Calculation\n",
+ "R0=R15/(1+15*a)\n",
+ "R50=R0*(1+a*50)\n",
+ "\n",
+ "#Result\n",
+ "print\"The value of resistance at 50 degree C is \", round(R50,2),\"ohm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of resistance at 50 degree C is 90.57 ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 154
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.21 Page no 271"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R20=20 #ohm\n",
+ "P=60 #W\n",
+ "V=120.0 #Volts\n",
+ "a=5*10**-3\n",
+ "\n",
+ "#Calculation\n",
+ "I=P/V\n",
+ "Rt=V/I\n",
+ "t=(((Rt/R20)-1)/a)+R20\n",
+ "\n",
+ "#Result\n",
+ "print\"Normal working temperature of the lamp is \", t,\"degree C\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Normal working temperature of the lamp is 2220.0 degree C\n"
+ ]
+ }
+ ],
+ "prompt_number": 160
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.22 Page no 271"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R0=5 #ohm\n",
+ "R100=5.23 #ohm\n",
+ "Rt=5.795 #ohm\n",
+ "\n",
+ "#Calculation\n",
+ "t=((Rt-R0)/(R100-R0))*100\n",
+ "\n",
+ "#Result\n",
+ "print\"The temperature of the bath is \", round(t,2),\"degree C\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The temperature of the bath is 345.65 degree C\n"
+ ]
+ }
+ ],
+ "prompt_number": 165
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.23 Page no 272"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "A=15*10**-4 #m**2\n",
+ "a=7.6*10**-8 # ohm m\n",
+ "l=2000 #m\n",
+ "b=0.005 #degree/C\n",
+ "\n",
+ "#Calculation\n",
+ "R0=a*l/A\n",
+ "R50=R0*(1+(b*50))\n",
+ "\n",
+ "#Result\n",
+ "print\"The value of resistance is \", round(R50,3),\"ohm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of resistance is 0.127 ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 172
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.24 Page no 272"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=0.004\n",
+ "ac=0.0007\n",
+ "R0=100\n",
+ "\n",
+ "#Calculation\n",
+ "R=ac*R0/a\n",
+ "\n",
+ "#Result\n",
+ "print\"The resistance of a copper filament is \", R,\"ohm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The resistance of a copper filament is 17.5 ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 175
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.28 Page no 281"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R1=4.0 #ohm\n",
+ "R2=4.0 #ohm\n",
+ "\n",
+ "#Calculation\n",
+ "Rab=1/((1/R1)+(1/R2))\n",
+ "\n",
+ "#Result\n",
+ "print\"The equivalent resisatance is \", Rab,\"ohm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The equivalent resisatance is 2.0 ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 178
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.29 Page no 281"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R1=15 #ohm\n",
+ "R2=30 #ohm\n",
+ "\n",
+ "#Calculation\n",
+ "R=R1*R2/(R1+R2)\n",
+ "\n",
+ "#Result\n",
+ "print\"The equivqlent resistance between A and B is \", R,\"ohm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The equivqlent resistance between A and B is 10 ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 181
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.31 Page no 282"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R1=5 #ohm\n",
+ "R2=9 #ohm\n",
+ "R3=14 #ohm\n",
+ "R4=11\n",
+ "R5=7\n",
+ "R6=18\n",
+ "R7=13\n",
+ "R8=22\n",
+ "V=22\n",
+ "\n",
+ "#Calculation\n",
+ "Rec=(R1+R2)*R3/(R1+R2+R3)\n",
+ "Rbe=(R4+R5)*R6/(R4+R5+R6)\n",
+ "Rae=(R7+R2)*R8/(R7+R2+R8)\n",
+ "I=V/Rae\n",
+ "\n",
+ "#Result\n",
+ "print\"The value of current in the branch AF is \", I,\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of current in the branch AF is 2 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 187
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.32 Page no 282"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R1=12 #ohm\n",
+ "R2=6 #ohm\n",
+ "\n",
+ "#Calculation\n",
+ "Rdg=R1*R2/(R1+R2)\n",
+ "Rch=R1*R2/(R1+R2)\n",
+ "Rab=Rdg+Rch\n",
+ "\n",
+ "#Result\n",
+ "print\"The equivalent resistance is \", Rab,\"ohm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The equivalent resistance is 8 ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 191
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.33 Page no 283"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Rab=500.0 #ohm\n",
+ "Rl=500 #ohm\n",
+ "Rbc=1500 #ohm\n",
+ "E=50 #Volts\n",
+ "Rac=2000.0 #ohm\n",
+ "V=40\n",
+ "\n",
+ "#Calculation\n",
+ "R=Rbc*Rl/(Rbc+Rl)\n",
+ "I=E/(Rab+R)\n",
+ "Pd=I*Rab\n",
+ "Rl1=E-Pd\n",
+ "I1=E/Rac\n",
+ "R12=V/I1\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Potential difference across the road is \", round(Rl1,2),\"V\"\n",
+ "print\"(ii) Resistance at BC is \", R12,\"ohm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Potential difference across the road is 21.43 V\n",
+ "(ii) Resistance at BC is 1600.0 ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 206
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.35 Page no 284"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R1=5 #ohm\n",
+ "R2=5.0 #ohm\n",
+ "R=6\n",
+ "\n",
+ "#Calculation\n",
+ "n=(1/(R-R1)*R2)\n",
+ "\n",
+ "#Result\n",
+ "print\"There are\", n,\"resistance are in parallel\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "There are 5.0 resistance are in parallel\n"
+ ]
+ }
+ ],
+ "prompt_number": 210
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.36 Page no 284"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R1=20.0 #ohm\n",
+ "R2=10.0 #ohm\n",
+ "R4=10\n",
+ "\n",
+ "#Calculation\n",
+ "Rbd=(R1*R2)/(R1+R2)\n",
+ "Rae=R2+Rbd+R4\n",
+ "\n",
+ "#Result\n",
+ "print\"The value of resistance is \", round(Rae,2),\"ohm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of resistance is 26.67 ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 218
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.37 Page no 284"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R1=2.0 #ohm\n",
+ "R2=3 #ohm\n",
+ "R3=2.8\n",
+ "E=6 #V\n",
+ "\n",
+ "#Calculation\n",
+ "Rab=R1*R2/(R1+R2)\n",
+ "Rt=Rab+R3\n",
+ "I=E/Rt\n",
+ "Vab=I*Rab\n",
+ "I1=Vab/2.0\n",
+ "\n",
+ "#Result\n",
+ "print\"The steady state current is \", I1,\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The steady state current is 0.9 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 226
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.38 Page no 285"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R1=3 #ohm\n",
+ "R2=3\n",
+ "R3=6\n",
+ "\n",
+ "#Calculation\n",
+ "Rad=(R1+R2)*R3/(R1+R2+R3)\n",
+ "Rae=(Rad+R1)*R3/(Rad+R1+R3)\n",
+ "Raf=(Rae+R1)*R3/(Rae+R1+R3)\n",
+ "Rab=(Raf+R1)*R2/(Rae+R1+R2)\n",
+ "\n",
+ "#Result\n",
+ "print\"the effective resistance between the point A and B is\", Rab,\"Ohm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the effective resistance between the point A and B is 2 Ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 234
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 39 Page no 285"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R2=50.0 #ohm\n",
+ "R3=50.0 #ohm\n",
+ "R4=75.0 #ohm\n",
+ "E=4.75\n",
+ "R1=100\n",
+ "\n",
+ "#Calculation\n",
+ "Rbc=1/((1/R2)+(1/R3)+(1/R4))\n",
+ "R=R1+Rbc\n",
+ "I=E/R\n",
+ "R11=I*R1\n",
+ "Vbc=E-(I*R1)\n",
+ "I2=Vbc/R2\n",
+ "I3=Vbc/R3\n",
+ "I4=Vbc/R4\n",
+ "\n",
+ "#Result\n",
+ "print\"Equivalent resistance of the circuit is \", R,\"ohm\"\n",
+ "print\"Current in R2 is\",I2,\"A\"\n",
+ "print\"Current in R3 is\",I3,\"A\"\n",
+ "print\"Current in R4 is\",I4,\"A\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Equivalent resistance of the circuit is 118.75 ohm\n",
+ "Current in R2 is 0.015 A\n",
+ "Current in R3 is 0.015 A\n",
+ "Current in R4 is 0.01 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 251
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.40 Page no 292"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "V=19\n",
+ "I1=0.5\n",
+ "I2=2 #A\n",
+ "r=2 \n",
+ "\n",
+ "#Calculation\n",
+ "E=V+I1*r\n",
+ "\n",
+ "#Result\n",
+ "print\"E.M.F is \", E,\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "E.M.F is 20.0 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 254
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.41 Page no 292"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "V=1.5\n",
+ "a=1.5\n",
+ "r1=0.5 #ohm\n",
+ "r2=0.25\n",
+ "R=2.25 #ohm\n",
+ "\n",
+ "#Calculation\n",
+ "E=V+a\n",
+ "r=r1+r2\n",
+ "Rt=r+R\n",
+ "I=E/Rt\n",
+ "pd=V-(I*r1)\n",
+ "pd1=V-(I*r2)\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) The circuit current is \",I,\"A\"\n",
+ "print\"(ii) P.D across the terminals of each cell is \",pd,\"V and \",pd1,\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) The circuit current is 1.0 A\n",
+ "(ii) P.D across the terminals of each cell is 1.0 V and 1.25 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 268
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.42 Page no 292"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "n=10\n",
+ "E=1.5\n",
+ "R=4 #ohm\n",
+ "r=0.1\n",
+ "a=8\n",
+ "\n",
+ "#Calculation\n",
+ "Emf=n*E\n",
+ "Rt=R+(n*r)\n",
+ "I=Emf/Rt\n",
+ "Emf1=(a*E)-(2*E)\n",
+ "I1=Emf1/Rt\n",
+ "I11=I-I1\n",
+ "\n",
+ "#Result\n",
+ "print\"Reduction in current is \", I11,\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Reduction in current is 1.2 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 277
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.43 Page no 292"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Emf=2\n",
+ "Emf1=1.9\n",
+ "Emf2=1.8\n",
+ "R1=0.05\n",
+ "R2=0.06\n",
+ "R3=0.07\n",
+ "R0=5 #ohm\n",
+ "\n",
+ "#Calculation\n",
+ "Emft=Emf+Emf1+Emf2\n",
+ "R=R1+R2+R3\n",
+ "Rt=R+R0\n",
+ "I=Emft/Rt\n",
+ "\n",
+ "#Result\n",
+ "print\"The reading of the ammeter is \", round(I,1),\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The reading of the ammeter is 1.1 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 283
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.44 Page no 293"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R1=6.0 #ohm\n",
+ "R2=3\n",
+ "I=0.8 #A\n",
+ "a=24\n",
+ "\n",
+ "#Calculation\n",
+ "I1=I*(R1+R2)/R1\n",
+ "I11=I1-I\n",
+ "Rp=R1*R2/(R1+R2)\n",
+ "Rt=Rp+8\n",
+ "r=(a/I1)-10\n",
+ "V=I1*Rt\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Current in 6 ohm resistance is \", I11,\"A\"\n",
+ "print\"(ii) Internal resistance of the battery is \", r,\"ohm\"\n",
+ "print\"(iii) The terminal potential difference of the battery is \", V,\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Current in 6 ohm resistance is 0.4 A\n",
+ "(ii) Internal resistance of the battery is 10.0 ohm\n",
+ "(iii) The terminal potential difference of the battery is 12.0 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 295
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.45 Page no 294"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R1=2 #ohm\n",
+ "R2=4\n",
+ "R3=6\n",
+ "E=8\n",
+ "r=1\n",
+ "\n",
+ "#Calculation\n",
+ "Rac=(R1+R2)*R3/(R1+R2+R3)\n",
+ "I=E/(Rac+r)\n",
+ "I1=I/2.0\n",
+ "\n",
+ "#Result\n",
+ "print\"Internal resistance is \", I1,\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Internal resistance is 1.0 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 299
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.46 Page no 294"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "E=1\n",
+ "R=2\n",
+ "\n",
+ "#Calculation\n",
+ "r=(E*R)-E\n",
+ "print\"The internal resisatnce of aech cell is \",r,\"ohm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The internal resisatnce of aech cell is 1 ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 18
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.47 Page no 294"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R1=15.0 # ohm\n",
+ "R2=15.0\n",
+ "E=2\n",
+ "V=1.6\n",
+ "\n",
+ "#Calculation\n",
+ "R=R1*R2/(R1+R2)\n",
+ "r=((E/V)-1)*R*4\n",
+ "\n",
+ "#Result\n",
+ "print\"Internal resisatnce of each cell is \", r,\"ohm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Internal resisatnce of each cell is 7.5 ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.48 Page no 295"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "I1=1 #A\n",
+ "E=1.5\n",
+ "I2=0.6\n",
+ "R2=2.33 #ohm\n",
+ "\n",
+ "#Calculation\n",
+ "R=2*E/I1\n",
+ "R1=2*E/I2\n",
+ "r=R1-2*R2\n",
+ "\n",
+ "#Result\n",
+ "print\"Internal resisatnce of each battery is \", r,\"ohm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Internal resisatnce of each battery is 0.34 ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 20
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.49 Page no 295"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R1=4 #ohm\n",
+ "R2=4 #ohm\n",
+ "R3=12\n",
+ "R4=6.0\n",
+ "E=16\n",
+ "r=1 #ohm\n",
+ "\n",
+ "#calculation\n",
+ "Rab=R1*R2/(R1+R2)\n",
+ "Rcd=R3*R4/(R3+R4)\n",
+ "R=Rab+Rcd+1\n",
+ "I=E/(R+r)\n",
+ "I1=I/2.0\n",
+ "I3=I*R4/(R3+R4)\n",
+ "I4=I*R3/(R3+R4)\n",
+ "Vab=4*I1\n",
+ "Vbc=I*1\n",
+ "Vcd=12*I3\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) equivalent resistance of the network is \", R,\"ohm\"\n",
+ "print\"(ii) Circuit current is\", I,\"A , Current in R1 is\",I1,\"A , Current in R3 is\",round(I3,2),\"A , Current in R4 is \",round(I4,2)\n",
+ "print \"Voltage drop Vab is\",Vab,\"V \\nVbc is\",Vbc,\"V \\nVcd is\",Vcd,\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) equivalent resistance of the network is 7.0 ohm\n",
+ "(ii) Circuit current is 2.0 A , Current in R1 is 1.0 A , Current in R3 is 0.67 A , Current in R4 is 1.33\n",
+ "Voltage drop Vab is 4.0 V \n",
+ "Vbc is 2.0 V \n",
+ "Vcd is 8.0 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 46
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter6_1.ipynb b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter6_1.ipynb new file mode 100644 index 00000000..f1c6f4b7 --- /dev/null +++ b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter6_1.ipynb @@ -0,0 +1,624 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:1bababaf98233cb133c1893aef0743afe5d48d8c1bcfe7a7487181ba9a8fde89"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 6 Electrical measurements"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.1 Page no 332"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=4\n",
+ "b=2.0\n",
+ "c=8\n",
+ "d=5\n",
+ "e=3.0\n",
+ "\n",
+ "#Calculation\n",
+ "I1=((a*c)+(b*e))/((b*c)+(d*e))\n",
+ "I2=(a-(2*I1))/e\n",
+ "V=(I1-I2)*5\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Current through each battery is\", round(I1,2),\"A and\",round(I2,2),\"A\"\n",
+ "print\"(ii) Terminal voltage is\",round(V,2),\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Current through each battery is 1.23 A and 0.52 A\n",
+ "(ii) Terminal voltage is 3.55 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 20
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.4 Page no 334"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=10\n",
+ "b=5.0\n",
+ "c=9.0\n",
+ "d=19.0\n",
+ "\n",
+ "#Calculation\n",
+ "I2=(a-c)/((b*a)-(d*c))\n",
+ "I1=(1-(5*I2))/c\n",
+ "I=I1+I2\n",
+ "pd=I*10\n",
+ "\n",
+ "#Result\n",
+ "print\"Current through each cell is\", round(I,2),\"A\"\n",
+ "print\"Potential difference across 10 ohm wire is\",round(pd,3),\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Current through each cell is 0.11 A\n",
+ "Potential difference across 10 ohm wire is 1.074 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 33
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.6 Page no 335"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=-3\n",
+ "b=4.0\n",
+ "c=3\n",
+ "\n",
+ "#Calculation\n",
+ "I1=a/(b+(c**2))\n",
+ "I2=-1-c*I1\n",
+ "I3=-(I1+I2)\n",
+ "\n",
+ "#Result\n",
+ "print\"Current through each cell is\", round(I1,2),\"A ,\",round(I2,2),\"A and\",round(I3,2),\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Current through each cell is -0.23 A , -0.31 A and 0.54 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 47
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.7 Page no 336"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=15\n",
+ "b=4\n",
+ "c=12.0\n",
+ "d=10\n",
+ "\n",
+ "#Calculation\n",
+ "R=(a*b)/c\n",
+ "X=(d*R)/(d-R)\n",
+ "\n",
+ "#Result\n",
+ "print\"The value of resistance is\", X,\"ohm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of resistance is 10.0 ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 51
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.8 Page no 336"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R1=4 #ohm\n",
+ "R2=3 #ohm\n",
+ "R3=2.0\n",
+ "R11=2.4 #ohm\n",
+ "E=6\n",
+ "\n",
+ "#Calculation\n",
+ "X=(R1*R2)/R3\n",
+ "R4=R2+X\n",
+ "R5=R1+R3\n",
+ "Rt=((R4*R5)/(R4+R5))+R11\n",
+ "I=E/Rt\n",
+ "\n",
+ "#Result\n",
+ "print\"the value of unknown resistance is\", X,\"ohm\"\n",
+ "print\"The current drawn by the circuit is\",I,\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the value of unknown resistance is 6.0 ohm\n",
+ "The current drawn by the circuit is 1.0 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 59
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.9 Page no 337"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=10\n",
+ "b=7.0\n",
+ "c=5\n",
+ "d=4\n",
+ "e=8.0\n",
+ "\n",
+ "#Calculation\n",
+ "I1=(a+a)/(b+1)\n",
+ "I3=(c+(4*I1))/e\n",
+ "I2=(-a+(6*I3)+I1)/2.0\n",
+ "\n",
+ "#Result\n",
+ "print\"Current I1=\",I1,\"A \\nI2=\",I2,\"A \\nI3=\",I3,\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Current I1= 2.5 A \n",
+ "I2= 1.875 A \n",
+ "I3= 1.875 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 70
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.11 Page no 338"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=28\n",
+ "b=5.0\n",
+ "c=2\n",
+ "\n",
+ "#Calculation\n",
+ "Rak=a/(b*c)\n",
+ "\n",
+ "#Result\n",
+ "print\"Total resistance from one end of vacant edge to other end is\", Rak,\"ohm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Total resistance from one end of vacant edge to other end is 2.8 ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 74
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.12 Page no 345"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R=10\n",
+ "l2=68.5\n",
+ "l1=58.3\n",
+ "\n",
+ "#Calculation\n",
+ "X=R*(l2/l1)\n",
+ "\n",
+ "#Result\n",
+ "print\"Value of X is\", round(X,1),\"ohm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Value of X is 11.7 ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 78
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.13 Page no 346"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R=2 #ohm\n",
+ "R1=2.4 #ohm\n",
+ "V=4 #V\n",
+ "E=1.5\n",
+ "\n",
+ "#Calculation\n",
+ "R11=R+R1\n",
+ "I=V/R11\n",
+ "Vab=I*R\n",
+ "K=Vab\n",
+ "l=E/K\n",
+ "\n",
+ "#Result\n",
+ "print\"Length for zero galvanometer deflection is\", l,\"m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Length for zero galvanometer deflection is 0.825 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 85
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.15 Page no 347"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "l1=33.7\n",
+ "l2=51.9\n",
+ "\n",
+ "#Calculation\n",
+ "S1=l1/(100-l1)\n",
+ "s11=l2/(100-l2)\n",
+ "s=((s11*12)/S1)-12\n",
+ "R=s*S1\n",
+ "\n",
+ "#Result\n",
+ "print\"Value of R is\", round(R,2),\"ohm \\nValue of S is\",round(s,1),\"ohm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Value of R is 6.85 ohm \n",
+ "Value of S is 13.5 ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 95
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.16 Page no 347"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=0.4\n",
+ "b=0.6\n",
+ "lab=10\n",
+ "\n",
+ "#Calculation\n",
+ "K=a/b\n",
+ "Vab=K*lab\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Potentila gradient along AB is\",round(K,2),\"V/m\"\n",
+ "print \"(ii) P.D between point A and B is\",round(Vab,2),\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Potentila gradient along AB is 0.67 V/m\n",
+ "(ii) P.D between point A and B is 6.67 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 103
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.17 Page no 348"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R1=990 #ohm\n",
+ "R=10.0 #ohm\n",
+ "E=2\n",
+ "l=1000 #mm\n",
+ "l1=400 #mm\n",
+ "\n",
+ "#Calculation\n",
+ "Rt=R1+R\n",
+ "I=E/Rt\n",
+ "pd=I*R\n",
+ "K=pd/l\n",
+ "pd1=K*l1\n",
+ "\n",
+ "#Result\n",
+ "print\"e.m.f. generated by the thermocouple is\", pd1,\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "e.m.f. generated by the thermocouple is 0.008 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 111
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.18 Page no 348"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "AB=600 #cm\n",
+ "AC=500.0 #cm\n",
+ "l=40*10**-3 #A\n",
+ "E=2\n",
+ "r=10\n",
+ "\n",
+ "#Calculation\n",
+ "R=2*AB/(AC*l)\n",
+ "K=2/AC\n",
+ "AC1=AC-r\n",
+ "pd=K*AC1\n",
+ "Iv=(E-pd)/r\n",
+ "R1=pd/Iv\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) The resistance of the whole wire is\", R,\"ohm\"\n",
+ "print\"(ii) Reading of voltmeter is\", pd,\"V\"\n",
+ "print\"(iii) Resistance of the voltmeter is\",R1,\"ohm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) The resistance of the whole wire is 60.0 ohm\n",
+ "(ii) Reading of voltmeter is 1.96 V\n",
+ "(iii) Resistance of the voltmeter is 490.0 ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 124
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.20 Page no 350"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=6\n",
+ "b=2\n",
+ "\n",
+ "#Calculation\n",
+ "R1=a/((b*b)-1)\n",
+ "R2=b*R1\n",
+ "\n",
+ "#Result\n",
+ "print\"Resistance R1 is\", R1,\"ohm \\nR2 is\",R2,\"Ohm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Resistance R1 is 2 ohm \n",
+ "R2 is 4 Ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 130
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.21 Page no 350"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R=20 #ohm\n",
+ "L=10 #m\n",
+ "pd=10**-3 #V/m\n",
+ "V=10**-2 #Volts\n",
+ "\n",
+ "#Calculation\n",
+ "I=V/R\n",
+ "R11=(2/I)-R\n",
+ "\n",
+ "#Result\n",
+ "print\"The value of resistance is\", R11,\"ohm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of resistance is 3980.0 ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 134
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter7_1.ipynb b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter7_1.ipynb new file mode 100644 index 00000000..74f3f441 --- /dev/null +++ b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter7_1.ipynb @@ -0,0 +1,735 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:47b2c0fcc74d4ba925e8938987dfe5c551c445c65c0145d8b28ae4df323cfc30"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 7 Heating effect of electric curent"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.1 Page no 373"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "V=240 #V\n",
+ "P=60\n",
+ "P1=100\n",
+ "\n",
+ "#Calculation\n",
+ "R=V**2/P\n",
+ "R1=V**2/P1\n",
+ "\n",
+ "#Result\n",
+ "print\"Resistance of a bulb for 60 W is\", R,\"ohm and for 100 W is\",R1,\"ohm\"\n",
+ "print\"Hence The resistance of 60 W,240 V bulb is more than that of 100 W,240 V bulb\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Resistance of a bulb for 60 W is 960 ohm and for 100 W is 576 ohm\n",
+ "Hence The resistance of 60 W,240 V bulb is more than that of 100 W,240 V bulb\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.2 Page no 373"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "V=230 #v\n",
+ "P=100\n",
+ "t=20*60\n",
+ "V1=115 #V\n",
+ "\n",
+ "#Calculation\n",
+ "R=V**2/P\n",
+ "E=(V1**2*t)/R\n",
+ "\n",
+ "#Result\n",
+ "print\"Heat and light energy is\", E,\"J\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Heat and light energy is 30000 J\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.3 Page no 374"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "P=500 #W\n",
+ "V=200.0 #V\n",
+ "V1=240\n",
+ "\n",
+ "#Calculation\n",
+ "I=P/V\n",
+ "R=V1-V\n",
+ "R1=R/I\n",
+ "\n",
+ "#Result\n",
+ "print\"The value of R=\",R1,\"ohm\"\n",
+ "print\"Current in a circuit is\",I,\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of R= 16.0 ohm\n",
+ "Current in a circuit is 2.5 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 18
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.4 Page no 374"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "P1=100.0 #W\n",
+ "P=1100.0 #W\n",
+ "V=250\n",
+ "\n",
+ "#Calculation\n",
+ "P2=P-P1\n",
+ "R=V**2/P2\n",
+ "\n",
+ "#Result\n",
+ "print\"The value of unknown resistance is\", R,\"ohm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of unknown resistance is 62.5 ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 25
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.6 Page no 374"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "V=220\n",
+ "P=200.0\n",
+ "P1=100\n",
+ "\n",
+ "#Calculation\n",
+ "R1=V**2/P\n",
+ "R2=V**2/P1\n",
+ "H=R1/R2\n",
+ "\n",
+ "#Result\n",
+ "print\"The ratio of heats genetated in them is\", H"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The ratio of heats genetated in them is 0.5\n"
+ ]
+ }
+ ],
+ "prompt_number": 29
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.7 Page no 375"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "m=1\n",
+ "c=1\n",
+ "a=100 #W\n",
+ "b=15\n",
+ "t=7.5 #second\n",
+ "P=1 #KW\n",
+ "C=860 #Kcal\n",
+ "\n",
+ "#Calculation\n",
+ "A=m*c*(a-b)\n",
+ "B=P*t/60.0\n",
+ "D=B*C\n",
+ "n=A*a/D\n",
+ "\n",
+ "#Result\n",
+ "print\"Efficiency of the kettle is\", round(n,1),\"%\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Efficiency of the kettle is 79.1 %\n"
+ ]
+ }
+ ],
+ "prompt_number": 36
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.8 Page no 375"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "P1=9 #W\n",
+ "R1=8\n",
+ "R2=12.0\n",
+ "\n",
+ "#Calculation\n",
+ "P2=(P1*R1)/R2\n",
+ "\n",
+ "#Result\n",
+ "print\"Power dissipated in 12 ohm resistor is\", P2,\"W\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Power dissipated in 12 ohm resistor is 6.0 W\n"
+ ]
+ }
+ ],
+ "prompt_number": 39
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.9 Page no 375"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "H1=10\n",
+ "a=5.0\n",
+ "b=4.2\n",
+ "\n",
+ "#Calculation\n",
+ "I1=(H1*b)/(a*4)\n",
+ "A=I1*4/b\n",
+ "\n",
+ "#Result\n",
+ "print\"Heat generated in 4 ohm resistor is\", A,\"cal/sec\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Heat generated in 4 ohm resistor is 2.0 cal/sec\n"
+ ]
+ }
+ ],
+ "prompt_number": 44
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.10 Page no 375"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "E=12 #V\n",
+ "I=1 #A\n",
+ "r=0.5 #ohm\n",
+ "\n",
+ "#Calculation\n",
+ "P1=E*I\n",
+ "P2=I**2*r\n",
+ "P=P1-P2\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Rate of consumption of chemical energy is\", P1,\"W\"\n",
+ "print\"(ii) Rate Of energy dissipated inside the battery is\",P2,\"W\"\n",
+ "print\"(iv) Rate of energy dissipated in the resistor is\", P,\"W\"\n",
+ "print\"(v) Power output of the source is\",P,\"W\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Rate of consumption of chemical energy is 12 W\n",
+ "(ii) Rate Of energy dissipated inside the battery is 0.5 W\n",
+ "(iv) Rate of energy dissipated in the resistor is 11.5 W\n",
+ "(v) Power output of the source is 11.5 W\n"
+ ]
+ }
+ ],
+ "prompt_number": 55
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.11 Page no 376"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "P=110 #W\n",
+ "P1=100 #W\n",
+ "n=5\n",
+ "V=220 #V\n",
+ "t=2 #hours\n",
+ "n1=4\n",
+ "P2=1120 #W\n",
+ "m=1.5 #per KWh\n",
+ "\n",
+ "#Calculation\n",
+ "W=n*P1\n",
+ "W1=V*t\n",
+ "W2=n1*P\n",
+ "W3=W+W1+W2+P2\n",
+ "E=(W3*t)*10**-3\n",
+ "E2=E*30\n",
+ "B=m*E2\n",
+ "\n",
+ "#Result\n",
+ "print\"Electricity bill for the month of september is\", B,\"Rs\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Electricity bill for the month of september is 225.0 Rs\n"
+ ]
+ }
+ ],
+ "prompt_number": 68
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.12 Page no 376"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "V=220 #V\n",
+ "P=60.0 #W\n",
+ "P1=85 #w\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "R=V**2/P\n",
+ "V1=math.sqrt(P1*R)\n",
+ "\n",
+ "#Result\n",
+ "print\"Maximum voltage is\", round(V1,1),\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Maximum voltage is 261.9 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 75
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.13 Page no 376"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "V=200 #V\n",
+ "P=500.0 #W\n",
+ "V1=160 #v\n",
+ "\n",
+ "#Calculation\n",
+ "R=V**2/P\n",
+ "H=V1**2/R\n",
+ "P1=P-H\n",
+ "H1=P1*100/P\n",
+ "\n",
+ "#Result\n",
+ "print\"Heat percentage is\", H1,\"%\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Heat percentage is 36.0 %\n"
+ ]
+ }
+ ],
+ "prompt_number": 82
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.14 Page no 376"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "P1=500 #W\n",
+ "P2=100\n",
+ "\n",
+ "#Calculation\n",
+ "R=P1/P2\n",
+ "\n",
+ "#Result\n",
+ "print\"Since P1'=5P2', 100W bulb will glow brighter\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Since P1'=5P2', 100W bulb will glow brighter\n"
+ ]
+ }
+ ],
+ "prompt_number": 86
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.15 Page no 377"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "m=900\n",
+ "w=100.0\n",
+ "c=1\n",
+ "a=80\n",
+ "b=4.2\n",
+ "V=210 #V\n",
+ "x=12\n",
+ "y=60\n",
+ "\n",
+ "#Calculation\n",
+ "Hout=(m+w)*c*a\n",
+ "Hin=(V*x*y)/b\n",
+ "Hin1=90/w*Hin\n",
+ "I=Hout/Hin1\n",
+ "\n",
+ "#Result\n",
+ "print\"Strength of the current is\", round(I,3),\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Strength of the current is 2.469 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 95
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.16 Page no 377"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=0.8\n",
+ "\n",
+ "#Calculation\n",
+ "H=a**2\n",
+ "H1=(1-H)*100\n",
+ "\n",
+ "#Result\n",
+ "print\"Decreased percentage is\", H1,\"%\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Decreased percentage is 36.0 %\n"
+ ]
+ }
+ ],
+ "prompt_number": 99
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.17 Page no 377"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=14\n",
+ "b=60\n",
+ "c=24\n",
+ "d=7.0\n",
+ "\n",
+ "#Calculation\n",
+ "t=a*b/60.0\n",
+ "t1=(c/d)\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Time in series is\", t,\"minute\"\n",
+ "print\"(ii) Time in parallel is\",round(t1,2),\"minute\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Time in series is 14.0 minute\n",
+ "(ii) Time in parallel is 3.43 minute\n"
+ ]
+ }
+ ],
+ "prompt_number": 108
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.19 Page no 378"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "I=0.5\n",
+ "R=100\n",
+ "t=30\n",
+ "a=4.2\n",
+ "m=200 #g\n",
+ "w=10 #g\n",
+ "\n",
+ "#Calculation\n",
+ "H=I**2*R*t*60/a\n",
+ "A=H/(m+w)\n",
+ "\n",
+ "#Result\n",
+ "print\"The rise of temperature is\", round(A,2),\"degree C\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The rise of temperature is 51.02 degree C\n"
+ ]
+ }
+ ],
+ "prompt_number": 113
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.20 Page no 379"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "c=4.2 #KJ/Kg/C\n",
+ "m=0.2 #Kg\n",
+ "a=90\n",
+ "b=20\n",
+ "t=30\n",
+ "V=230\n",
+ "\n",
+ "#calculation\n",
+ "d=a-b\n",
+ "H=c*m*d\n",
+ "P=H/t\n",
+ "I=P/V\n",
+ "\n",
+ "#Result\n",
+ "print\"The value of current is\", round(I*10**3,2),\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of current is 8.52 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 120
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter8_1.ipynb b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter8_1.ipynb new file mode 100644 index 00000000..baadbaea --- /dev/null +++ b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter8_1.ipynb @@ -0,0 +1,643 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:db58e543ef86cd601814ac49a8404db7a1403e7140977a41ff4c6b1fc2ae61b9"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 8 Magnetic field due to electric current"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 8.1 Page no 417"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "q=1.6*10**-19 #c\n",
+ "B=0.1 #T\n",
+ "v=5.0*10**6 #m/s\n",
+ "a=90 #degree\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "Fm=q*v*B*math.sin(a)\n",
+ "\n",
+ "#Result\n",
+ "print\"Force on the proton is\", round(Fm*10**14,1)*10**-14,\"N\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Force on the proton is 7.2e-14 N\n"
+ ]
+ }
+ ],
+ "prompt_number": 36
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 8.2 Page no 417"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "n=1.0*10**29 #m**-3\n",
+ "e=1.6*10**-19 #C\n",
+ "A=2*10**-6 #m**2\n",
+ "I=5 #A\n",
+ "B=0.15 #T\n",
+ "a=90 #degree\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "Vd=I/(n*e*A)\n",
+ "Fm=e*Vd*B*math.sin(a)\n",
+ "\n",
+ "#Result\n",
+ "print\"Force acting on each electron is\", round(Fm*10**24,2)*10**-24,\"N\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Force acting on each electron is 3.35e-24 N\n"
+ ]
+ }
+ ],
+ "prompt_number": 34
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 8.3 Page no 417"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "q=2*1.6*10**-19 #C\n",
+ "v=6*10**5 #m/s\n",
+ "B=0.2 #T\n",
+ "a=90 #degree\n",
+ "m=6.65*10**-27\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "Fm=q*v*B*math.sin(a)\n",
+ "a=Fm/m\n",
+ "\n",
+ "#Result\n",
+ "print\"Force on alpha particle is\", round(Fm*10**14,2)*10**-14,\"N\"\n",
+ "print\"Acceleration of alpha particle is\",round(a*10**-12,2)*10**12,\"m/s**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Force on alpha particle is 3.43e-14 N\n",
+ "Acceleration of alpha particle is 5.16e+12 m/s**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 44
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 8.4 Page no 426"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=60 #degree\n",
+ "u=4*3.14*10**-7 #T/A m\n",
+ "Bc=2\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "a=(Bc/2.0)/(math.tan(60)*180/3.14)\n",
+ "B1=(10**-7*math.tan(60)*(math.sin(60*180/3.14)+math.sin(60*180/3.14)))*10\n",
+ "B=3*B1\n",
+ "\n",
+ "#Result\n",
+ "print\"Magnetic fieldat the centroid of the triangle is\", round(B*10**7,0),\"*10**-7 T\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Magnetic fieldat the centroid of the triangle is 10.0 *10**-7 T\n"
+ ]
+ }
+ ],
+ "prompt_number": 64
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 8.5 Page no 426"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "n=20\n",
+ "I=1 #A\n",
+ "r=0.08 #m\n",
+ "u=4*3.14*10**-7 #T/A m\n",
+ "\n",
+ "#Calculation\n",
+ "B=u*n*I/(2*r)\n",
+ "\n",
+ "#Result\n",
+ "print\"Magnitude of the magnetic field is\", B*10**4,\"*10*4 T\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Magnitude of the magnetic field is 1.57 *10*4 T\n"
+ ]
+ }
+ ],
+ "prompt_number": 68
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 8.6 Page no 426"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "u=10**-7\n",
+ "I=10*10**-2 #A\n",
+ "r=0.5\n",
+ "\n",
+ "#Calculation\n",
+ "B=u*I/r**2\n",
+ "\n",
+ "#Result\n",
+ "print\"Magnetic field on Y axis is\", B,\"K^ T\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Magnetic field on Y axis is 4e-08 K^ T\n"
+ ]
+ }
+ ],
+ "prompt_number": 71
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 8.7 Page no 426"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "I=5 #A\n",
+ "l=0.01 #m\n",
+ "a=45 #degree\n",
+ "r=2 #m\n",
+ "u=10**-7\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "B=(u*I*l*math.sin(a)*180/3.14)/r**2\n",
+ "\n",
+ "#Result\n",
+ "print\"Magnetic field is\", round(B*10**8,1)*10**-10,\"T\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Magnetic field is 6.1e-10 T\n"
+ ]
+ }
+ ],
+ "prompt_number": 82
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 8.8 Page no 427"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "u=4*3.14*10**-7 #T/A m\n",
+ "n=20\n",
+ "I=12 #A\n",
+ "r=0.1 #m\n",
+ "\n",
+ "#Calculation\n",
+ "B=u*n*I/(2*r)\n",
+ "\n",
+ "#Result\n",
+ "print\"Magnetic field at the centre of coil is\", round(B*10**3,1),\"*10**-3 T\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Magnetic field at the centre of coil is 1.5 *10**-3 T\n"
+ ]
+ }
+ ],
+ "prompt_number": 87
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 8.12 Page no 429"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "r=0.02 #m\n",
+ "\n",
+ "#Calculation\n",
+ "B=u*I/(4*r)\n",
+ "\n",
+ "#Result\n",
+ "print\"The magnitude of magnetic field is\", round(B*10**4,2),\"*10**-4 T\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The magnitude of magnetic field is 1.88 *10**-4 T\n"
+ ]
+ }
+ ],
+ "prompt_number": 92
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 8.13 Page no 429"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "v=4*10**6\n",
+ "r=0.5*10**-10\n",
+ "e=1.6*10**-19\n",
+ "t=1\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "f=v/(2.0*math.pi*r)\n",
+ "I=f*e/t\n",
+ "B=u*I/(2*r)\n",
+ "\n",
+ "#Result\n",
+ "print\"Magnetic field produced by the electrons is\", round(B,1),\"T\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Magnetic field produced by the electrons is 25.6 T\n"
+ ]
+ }
+ ],
+ "prompt_number": 98
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 8.15 Page no 430"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "n=100\n",
+ "I=5 #A\n",
+ "r=0.1 #m\n",
+ "x=0.05\n",
+ "\n",
+ "#Calculation\n",
+ "B=u*n*I/(2*r)\n",
+ "B1=(u*n*I*r**2)/(2.0*(r**2+x**2)**1.5)\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Magnetic field at the centre of the coil is\",B*10**3,\"*10**-3 T\"\n",
+ "print\"(ii) The magnetic field at the point on the axis of the coil is\",round(B1*10**3,2),\"*10**-3 T\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Magnetic field at the centre of the coil is 3.14 *10**-3 T\n",
+ "(ii) The magnetic field at the point on the axis of the coil is 2.25 *10**-3 T\n"
+ ]
+ }
+ ],
+ "prompt_number": 109
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 8.18 Page no 431"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=5*10**-2\n",
+ "I=50\n",
+ "e=1.6*10**-19\n",
+ "B1=10**7\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "B=u*I/(2*math.pi*a)\n",
+ "F=e*B1*B\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Force on electron when velocity is towards the wire\", round(F*10**16,1)*10**-16,\"N\"\n",
+ "print\"(ii) Force on electron when velocity is parallel to the wire\", round(F*10**16,1)*10**-16,\"N\"\n",
+ "print\"(iii) Force on electron when velocity is perpendicular to the wire is zero\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Force on electron when velocity is towards the wire 3.2e-16 N\n",
+ "(ii) Force on electron when velocity is parallel to the wire 3.2e-16 N\n",
+ "(iii) Force on electron when velocity is perpendicular to the wire is zero\n"
+ ]
+ }
+ ],
+ "prompt_number": 118
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 8.20 Page no 432"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "e=1.6*10**-19\n",
+ "f=6.8*10**15\n",
+ "r=0.51*10**-10\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "I=e*f\n",
+ "B=(u*I)/(2*r)\n",
+ "M=1*I*math.pi*r**2\n",
+ "\n",
+ "#Result\n",
+ "print\"The effective dipole moment is\",round(M*10**24,0)*10**-24,\"Am**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The effective dipole moment is 9e-24 Am**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 127
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 8.22 Page no 439"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "n=5*850/1.23\n",
+ "I=5.57 #A\n",
+ "\n",
+ "#calculation\n",
+ "u=4*math.pi*10**-7\n",
+ "B=u*n*I\n",
+ "\n",
+ "#Result\n",
+ "print\"Magnitude of magnetic field is\", round(B*10**3,1),\"*10**-3 T\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Magnitude of magnetic field is 24.2 *10**-3 T\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 8.23 Page no 439"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "r1=20\n",
+ "r2=25\n",
+ "I=2 #a\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "r=(r1+r2)/2.0\n",
+ "l=(2*math.pi*r)*10**-2\n",
+ "n=1500/l\n",
+ "B=u*n*I\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Magnetic field inside the toroid is\", round(B,3),\"T\"\n",
+ "print\"(ii) magnetic field outside the toroid is zero\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Magnetic field inside the toroid is 0.003 T\n",
+ "(ii) magnetic field outside the toroid is zero\n"
+ ]
+ }
+ ],
+ "prompt_number": 138
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 8.25 Page no 440"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "I=2 #A\n",
+ "R=5*10**-2 #m\n",
+ "r=3*10**-2 #m\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "B=u*I*r/(2*math.pi*R**2)\n",
+ "\n",
+ "#Result\n",
+ "print round(B*10**6,1),\"*10**-6 T\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "4.8 *10**-6 T\n"
+ ]
+ }
+ ],
+ "prompt_number": 142
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter9_1.ipynb b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter9_1.ipynb new file mode 100644 index 00000000..7669d0e6 --- /dev/null +++ b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter9_1.ipynb @@ -0,0 +1,1740 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:02dc05916beb2a686e89acb729415599e9656d95fc169884e1d4a92b0e8ee888"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 9 Motion of charged particles in electric and magnetic motion"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.1 Page no 472"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "V=90 #V\n",
+ "d=2.0*10**-2\n",
+ "e=1.8*10**11\n",
+ "x=5*10**-2\n",
+ "v=10**7\n",
+ "\n",
+ "#Calculation\n",
+ "E=V/d\n",
+ "a=e*E\n",
+ "t=x/v\n",
+ "y=0.5*a*t**2\n",
+ "\n",
+ "#Result\n",
+ "print\"Transverse deflection produced by electric field is\", round(y*10**2,1),\"cm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Transverse deflection produced by electric field is 1.0 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.2 Page no 473"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "V=500\n",
+ "d=2*10**-2 #m\n",
+ "v=3*10**7\n",
+ "x=6*10**-2\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "E=V/d\n",
+ "a=E*e\n",
+ "t=x/v\n",
+ "v1=a*t\n",
+ "T=v1/v\n",
+ "A=math.atan(T)*180.0/3.14\n",
+ "\n",
+ "#Result\n",
+ "print\"Angle is\", round(A,1),\"degree\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Angle is 16.7 degree\n"
+ ]
+ }
+ ],
+ "prompt_number": 31
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.3 Page no 474"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "x=10*10**-2\n",
+ "v=3*10**7\n",
+ "S=1.76*10**-3\n",
+ "a=1800\n",
+ "\n",
+ "#Calculation\n",
+ "t=x/v\n",
+ "e=S*2/(a*t**2)\n",
+ "\n",
+ "#Result\n",
+ "print\"Specific charge of the electron is\", e,\"C/Kg\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Specific charge of the electron is 1.76e+11\n"
+ ]
+ }
+ ],
+ "prompt_number": 34
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.4 Page no 478"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "m=9*10**-31\n",
+ "v=3*10**7\n",
+ "q=1.6*10**-19 #C\n",
+ "B=6*10**-4\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "r=m*v/(q*B)\n",
+ "f=q*B/(2.0*math.pi*m)\n",
+ "E=(0.5*m*v**2)/1.6*10**-16\n",
+ "\n",
+ "#Result\n",
+ "print\"Energy is\", round(E*10**32,2),\"Kev\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Energy is 2.53 Kev\n"
+ ]
+ }
+ ],
+ "prompt_number": 44
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.5 Page no 479"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "m=9*10**-31\n",
+ "e=1.6*10**-19\n",
+ "V=100\n",
+ "B=0.004\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "r=math.sqrt(2*m*e*V)/(e*B)\n",
+ "\n",
+ "#Result\n",
+ "print\"Radius of the path is\", round(r*10**3,1),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Radius of the path is 8.4 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 50
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.6 Page no 479"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "m=1.67*10**-27\n",
+ "v=4*10**5\n",
+ "a=60\n",
+ "q=1.6*10**-19\n",
+ "B=0.3\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "r=(m*v*math.sin(a*3.14/180.0))/q*B\n",
+ "P=v*math.cos(a*3.14/180.0)*((2*math.pi*m)/(q*B))\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Radius of the helical path is\",round(r*10**3,1),\"cm\"\n",
+ "print\"(ii) Pitch of helix is\", round(P*10**2,2),\"cm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Radius of the helical path is 1.1 cm\n",
+ "(ii) Pitch of helix is 4.38 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 70
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.7 Page no 479"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "M=5*10**6 #ev\n",
+ "e=1.6*10**-19\n",
+ "m=1.6*10**-27\n",
+ "B=1.5\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "v=math.sqrt((2*M*e)/m)\n",
+ "F=q*v*B*math.sin(90*3.14/180.0)\n",
+ "\n",
+ "#Result\n",
+ "print\"Magnitude of the force is\", round(F*10**12,2)*10**-12,\"N\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Magnitude of the force is 7.59e-12 N\n"
+ ]
+ }
+ ],
+ "prompt_number": 78
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.8 Page no 480"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "m=1.67*10**-27 #Kg\n",
+ "v=4*10**5\n",
+ "B=0.3 #T\n",
+ "q=1.6*10**-19 #C\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "r=m*v*math.sin(60*3.14/180.0)/(q*B)\n",
+ "P=2*math.pi*r*1/(math.tan(60*3.14/180.0))\n",
+ "\n",
+ "#Result\n",
+ "print\"Pitch of the helix is\", round(P*10**2,2),\"cm\"\n",
+ "print\"Radius of helical path is\",round(r*10**2,3),\"cm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Pitch of the helix is 4.38 cm\n",
+ "Radius of helical path is 1.205 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.9 Page no 480"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "q=3.2*10**-19\n",
+ "B=1.2\n",
+ "r=0.45\n",
+ "m=6.8*10**-27\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "v=(q*B*r)/m\n",
+ "f=v/(2.0*math.pi*r)\n",
+ "K=(0.5*m*v**2)/(1.6*10**-19)\n",
+ "V=K/2.0\n",
+ "\n",
+ "#Result\n",
+ "print\"Required potentila difference is\", round(V*10**-6,0),\"*10**6 V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Required potentila difference is 7.0 *10**6 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 15
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.10 Page no 480"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "I=4\n",
+ "u=10**-7\n",
+ "a=0.2 #m\n",
+ "v=4*10**6\n",
+ "q=1.6*10**-19\n",
+ "\n",
+ "#Calculation\n",
+ "B=(u*2*I)/a\n",
+ "F=q*v*B\n",
+ "\n",
+ "#Result\n",
+ "print\"Force is\", F,\"N\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Force is 2.56e-18 N\n"
+ ]
+ }
+ ],
+ "prompt_number": 20
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.11 Page no 481"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "e=1.6*10**-19\n",
+ "a=10**6\n",
+ "\n",
+ "#Calculation\n",
+ "q=2*e\n",
+ "F=q*a\n",
+ "\n",
+ "#Result\n",
+ "print\"Magnitude force acting on the particle is\", F"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Magnitude force acting on the particle is 3.2e-13\n"
+ ]
+ }
+ ],
+ "prompt_number": 22
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.13 Page no 482"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "E=3.4*10**4 #V/m\n",
+ "B=2*10**-3 #Wb/m**2\n",
+ "m=9.1*10**-31\n",
+ "e=1.6*10**-19\n",
+ "\n",
+ "#Calculation\n",
+ "v=E/B\n",
+ "r=(m*v)/(e*B)\n",
+ "\n",
+ "#Result\n",
+ "print\"Radius of the circular path is\", round(r*10**2,1),\"*10**-2 m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Radius of the circular path is 4.8 *10**-2 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 29
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.14 Page no 482"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "V=600 #V\n",
+ "d=3*10**-3 #m\n",
+ "v=2*10**6 #m/s\n",
+ "\n",
+ "#Calculation\n",
+ "B=V/(d*v)\n",
+ "\n",
+ "#Result\n",
+ "print\"Magnitude of magnetic field is\", B,\"T\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Magnitude of magnetic field is 0.1 T\n"
+ ]
+ }
+ ],
+ "prompt_number": 32
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.15 Page no 487"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "q=1.6*10**-19 #c\n",
+ "B=2 #T\n",
+ "m=1.66*10**-27 #Kg\n",
+ "K=5*10**6\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "f=(q*B)/(2.0*math.pi*m)\n",
+ "v=math.sqrt((2*K*q)/m)\n",
+ "r=(m*v)/(q*B)\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) The frequency needed for applied alternating voltage is\", round(f*10**-7,0),\"*10**7 HZ\"\n",
+ "print\"(ii) Radius of the cyclotron is\",round(r,2),\"m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) The frequency needed for applied alternating voltage is 3.0 *10**7 HZ\n",
+ "(ii) Radius of the cyclotron is 0.16 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 45
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.16 Page no 487"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "B=1.7 #T\n",
+ "q=1.6*10**-19 #c\n",
+ "r=0.5\n",
+ "m=1.66*10**-27\n",
+ "\n",
+ "#Calculation\n",
+ "K=((B**2*q**2*r**2)/(2.0*m))/q\n",
+ "\n",
+ "#Result\n",
+ "print\"Kinetic energy of proton is\", round(K*10**-6,0),\"Mev\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Kinetic energy of proton is 35.0 Mev\n"
+ ]
+ }
+ ],
+ "prompt_number": 51
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.17 Page no 487"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "B=0.8\n",
+ "q=3.2*10**-19 #C\n",
+ "d=1.2\n",
+ "m=4*1.66*10**-27 #Kg\n",
+ "a=1.60*10**-19\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "r=d/2.0\n",
+ "K=(B**2*q**2*r**2)/(2.0*m*a)\n",
+ "v=(q*B*r)/m\n",
+ "f=(q*B)/(2.0*math.pi*m)\n",
+ "\n",
+ "#Result\n",
+ "print\"Frequency of alternating voltage is\", round(f*10**-7,2),\"*10**7 HZ\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Frequency of alternating voltage is 0.61 *10**7 HZ\n"
+ ]
+ }
+ ],
+ "prompt_number": 61
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.18 Page no 488"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "q=1.6*10**-19 #C\n",
+ "r=0.6 #m\n",
+ "m=1.67*10**-27 #Kg\n",
+ "f=10**7\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "B=(2*math.pi*m*f)/q\n",
+ "K=((B**2*q**2*r**2)/(2.0*m))/1.6*10**-13\n",
+ "\n",
+ "#Result\n",
+ "print\"Kinetic energy of the protons is\", round(K*10**26,1),\"Mev\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Kinetic energy of the protons is 7.4 Mev\n"
+ ]
+ }
+ ],
+ "prompt_number": 67
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.19 Page no 493"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "I=5 #A\n",
+ "l=0.06 #m\n",
+ "B=0.02 #T\n",
+ "a=90\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "F=I*B*l*math.sin(a*3.14/180.0)\n",
+ "\n",
+ "#Result\n",
+ "print\"Force is\", round(F,3),\"N\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Force is 0.006 N\n"
+ ]
+ }
+ ],
+ "prompt_number": 71
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.20 Page no 494"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "m=0.2 #Kg\n",
+ "I=2 #A\n",
+ "l=1.5 #m\n",
+ "g=9.8\n",
+ "\n",
+ "#Calculation\n",
+ "B=(m*g)/(I*l)\n",
+ "\n",
+ "#Result\n",
+ "print\"Magnitude of the magnetic field is\", round(B,2),\"T\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Magnitude of the magnetic field is 0.65 T\n"
+ ]
+ }
+ ],
+ "prompt_number": 76
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.21 Page no 494"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#given\n",
+ "r=0.002 #m\n",
+ "m=0.05\n",
+ "g=9.8\n",
+ "\n",
+ "#Calculation\n",
+ "u=4*math.pi*10**-7\n",
+ "f=u/(2*math.pi*r)\n",
+ "f1=m*g\n",
+ "I=math.sqrt(f1*f**-1)\n",
+ "\n",
+ "#Result\n",
+ "print\"Current in each wire is\", I,\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Current in each wire is 70.0 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 82
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.22 Page no 494"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "r=0.04 #m\n",
+ "I1=20\n",
+ "I2=16\n",
+ "l=0.15\n",
+ "r1=0.1\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "u=4*math.pi*10**-7\n",
+ "F1=(u*I1*I2*l)/(2.0*math.pi*r)\n",
+ "F2=(u*I1*I2*l)/(2.0*math.pi*r1)\n",
+ "F=F1-F2\n",
+ "\n",
+ "#Result\n",
+ "print\"Net force on the loop is\", F*10**4,\"*10**-4 N\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Net force on the loop is 1.44 *10**-4 N\n"
+ ]
+ }
+ ],
+ "prompt_number": 89
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.23 Page no 495"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "m=0.3 #Kg\n",
+ "a=30 #degree\n",
+ "B=0.15 #T\n",
+ "g=9.8 #m/s**2\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "I=(m*g*math.tan(a*3.14/180.0))/B\n",
+ "\n",
+ "#Result\n",
+ "print\"value of current is\", round(I,2),\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "value of current is 11.31 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 94
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.24 Page no 495"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "B=3*10**-5 #T\n",
+ "I=1 #A\n",
+ "\n",
+ "#Calculation\n",
+ "F=I*B*math.sin(90)\n",
+ "\n",
+ "#Result\n",
+ "print\"The direction of the force is downward i.e\", round(F*10**5,0),\"*10**-5 N/m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The direction of the force is downward i.e 3.0 *10**-5 N/m\n"
+ ]
+ }
+ ],
+ "prompt_number": 99
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "Example 9.25 Page no 495"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": []
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "m=1.2*10**-3\n",
+ "B=0.6 #T\n",
+ "g=9.8 #m/s**2\n",
+ "r=0.05\n",
+ "b=3.8\n",
+ "\n",
+ "#Calculation\n",
+ "I=(m*g)/B\n",
+ "R=r*b\n",
+ "V=I*R\n",
+ "\n",
+ "#Result\n",
+ "print\"Potentila difference is\", round(V*10**3,1),\"*10**-3 V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Potentila difference is 3.7 *10**-3 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 105
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.26 Page no 496"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "I2=10 #A\n",
+ "r=0.1 #m\n",
+ "l=2 #m\n",
+ "I1=2\n",
+ "I2=10\n",
+ "r=0.1\n",
+ "\n",
+ "#Calculation\n",
+ "u=4*math.pi*10**-7\n",
+ "F=u*I1*I2*I1/(2.0*math.pi*r)\n",
+ "\n",
+ "#Result\n",
+ "print\"Force on small conductor\", F,\"N\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Force on small conductor 8e-05 N\n"
+ ]
+ }
+ ],
+ "prompt_number": 109
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.27 Page no 500"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "A=10**-3 #m**\n",
+ "n=10\n",
+ "I=2 #A\n",
+ "B=0.1 #T\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "t=n*I*A*B*math.cos(0)\n",
+ "t1=n*I*A*B*math.cos(60*3.14/180.0)\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Torque when magnetic field is parallel to the field\", round(t*10**3,0),\"*10**-3 Nm\"\n",
+ "print\"(ii) Torque when magnetic field is perpendicular to the field is zero\"\n",
+ "print\"(iii) Torque when magnetic field is 60 degree to the field is\",round(t1*10**3,1),\"*10**-3 Nm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Torque when magnetic field is parallel to the field 2.0 *10**-3 Nm\n",
+ "(ii) Torque when magnetic field is perpendicular to the field is zero\n",
+ "(iii) Torque when magnetic field is 60 degree to the field is 1.0 *10**-3 Nm\n"
+ ]
+ }
+ ],
+ "prompt_number": 121
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.28 Page no 500"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "r=7\n",
+ "I=10\n",
+ "B=100*10**-4\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "A=math.pi*r**2\n",
+ "t=I*A*B\n",
+ "\n",
+ "#Result\n",
+ "print\"Magnitude of maximum torque is\", round(t*10**-1,2),\"*10**-3 Nm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Magnitude of maximum torque is 1.54 *10**-3 Nm\n"
+ ]
+ }
+ ],
+ "prompt_number": 127
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.29 Page no 501"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "N=10\n",
+ "I=0.06\n",
+ "r=0.05\n",
+ "n=1000\n",
+ "I2=25\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "A=math.pi*r**2\n",
+ "M=N*I*A\n",
+ "u=4*math.pi*10**-7\n",
+ "B=u*n*I2\n",
+ "t=M*B*math.sin(45*3.14/180.0)\n",
+ "\n",
+ "#Result\n",
+ "print\"Torgue is\", round(t*10**4,2),\"*10**-4 Nm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Torgue is 1.05 *10**-4 Nm\n"
+ ]
+ }
+ ],
+ "prompt_number": 134
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.30 Page no 501"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "n=100\n",
+ "l=3.2 \n",
+ "r=0.1\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "u=4*math.pi*10**-7\n",
+ "B=(u*n*l)/(2.0*r)\n",
+ "M=n*l*math.pi*r**2\n",
+ "t=M*B*math.sin(0)\n",
+ "t1=(M*B*math.sin(90*3.14/180.0))*10**3\n",
+ "w=math.sqrt((2*M*B*10**3)/r)\n",
+ "\n",
+ "#Result\n",
+ "print\"(a) Field at the centre of the coil is\", round(B*10**3,0),\"*10**-3 T\"\n",
+ "print\"(b) Magnetic moment of the coil is\",round(M,0),\"Am**2\"\n",
+ "print\"(c) Magnitude of the torque on the coil in the initial position is\",t\n",
+ "print\" Magnitude of the torque on the coil in the final position is\",round(t1,0),\"Nm\"\n",
+ "print \"(d) Angular speed acquired by the coil is\",round(w,0),\"rad/s\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) Field at the centre of the coil is 2.0 *10**-3 T\n",
+ "(b) Magnetic moment of the coil is 10.0 Am**2\n",
+ "(c) Magnitude of the torque on the coil in the initial position is 0.0\n",
+ " Magnitude of the torque on the coil in the final position is 20.0 Nm\n",
+ "(d) Angular speed acquired by the coil is 20.0 rad/s\n"
+ ]
+ }
+ ],
+ "prompt_number": 25
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.31 Page no 505"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "n=125\n",
+ "I=20*10**-3 #A\n",
+ "B=0.5 #T\n",
+ "A=400*10**-6 #m**2\n",
+ "K=40*10**-6\n",
+ "\n",
+ "#Calculation\n",
+ "T=n*I*B*A\n",
+ "a=T/K\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Torque exerted is\", T*10**4,\"*10**-4 Nm\"\n",
+ "print\"(ii) Angular deflection of the coil is\", a,\"degree\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Torque exerted is 5.0 *10**-4 Nm\n",
+ "(ii) Angular deflection of the coil is 12.5 degree\n"
+ ]
+ }
+ ],
+ "prompt_number": 32
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.32 Page no 505"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "K=3*10**-9 #Nm/deg\n",
+ "a=36\n",
+ "n=60\n",
+ "B=9*10**-3 #T\n",
+ "A=5*10**-5 #m**2\n",
+ "\n",
+ "#Calculation\n",
+ "I=(K*a)/(n*B*A)\n",
+ "\n",
+ "#Result\n",
+ "print\"Maximum current is\", I*10**3,\"mA\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Maximum current is 4.0 mA\n"
+ ]
+ }
+ ],
+ "prompt_number": 36
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.33 Page no 506"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "n=30\n",
+ "B=0.25 #T\n",
+ "A=1.5*10**-3\n",
+ "K=10**-3\n",
+ "\n",
+ "#Calculation\n",
+ "S=(n*B*A)/K\n",
+ "\n",
+ "#Result\n",
+ "print\"Current sensitivity of the galvanometer is\", S,\"degree/A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Current sensitivity of the galvanometer is 11.25 degree/A\n"
+ ]
+ }
+ ],
+ "prompt_number": 39
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.35 Page no 509"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Ig=0.015 #A\n",
+ "G=5\n",
+ "I=1\n",
+ "V=15\n",
+ "\n",
+ "#Calculation\n",
+ "S=(Ig*G)/(I-Ig)\n",
+ "R=G*S/(G+S)\n",
+ "R1=(V/Ig)-G\n",
+ "R2=R1+G\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Resistance of ammeter of range 0-1 A is\", R,\"ohm\"\n",
+ "print\"(ii) Resistance of ammeter of range 0-15 A is\", R2,\"ohm\" "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Resistance of ammeter of range 0-1 A is 0.075 ohm\n",
+ "(ii) Resistance of ammeter of range 0-15 A is 1000.0 ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 47
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.36 Page no 510"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "V=75 #mV\n",
+ "Ig=0.025 #A\n",
+ "I=25 #mA\n",
+ "I1=100\n",
+ "V1=750\n",
+ "\n",
+ "#Calculation\n",
+ "G=V/I\n",
+ "S=(Ig*G)/(I1-Ig)\n",
+ "R=(V1/Ig)-G\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Resistance for an ammeter of range 0-100 A is\", round(S,5),\"ohm\"\n",
+ "print\"(ii) Resistance for an ammeter of range 0-750 A is\", round(R,5),\"ohm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Resistance for an ammeter of range 0-100 A is 0.00075 ohm\n",
+ "(ii) Resistance for an ammeter of range 0-750 A is 29997.0 ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 58
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.37 Page no 510"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Rg=60\n",
+ "R1=3.0\n",
+ "rs=0.02\n",
+ "\n",
+ "#Calculation\n",
+ "Rt=Rg+R1\n",
+ "I=R1/Rt\n",
+ "Rm=(Rg*rs)/(Rg+rs)\n",
+ "R2=Rm+R1\n",
+ "I1=R1/R2\n",
+ "I2=R1/R1\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) The value of current is\", round(I,3),\"A\"\n",
+ "print\"(ii) The value of current is\", round(I1,2),\"A\"\n",
+ "print\"(iii) The value of current is\",I2,\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) The value of current is 0.048 A\n",
+ "(ii) The value of current is 0.99 A\n",
+ "(iii) The value of current is 1.0 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 74
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.38 Page no 511"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "V=100\n",
+ "v=1\n",
+ "a=1980\n",
+ "\n",
+ "#Calculation\n",
+ "Rm=a/(V-v)\n",
+ "\n",
+ "#Result\n",
+ "print\"Resistance of the voltmeter is\", Rm,\"ohm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Resistance of the voltmeter is 20 ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 77
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.39 Page no 511"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R1=1200.0 #ohm\n",
+ "R2=600 #ohm\n",
+ "Vab=5 #V\n",
+ "V=35\n",
+ "\n",
+ "#Calculation\n",
+ "Rp=(R1*R2)/(R1+R2)\n",
+ "I=Vab/Rp\n",
+ "pd=V-Vab\n",
+ "R=pd/I\n",
+ "\n",
+ "#Result\n",
+ "print\"value of unknown resistance is\", R,\"ohm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "value of unknown resistance is 2400.0 ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 84
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.40 Page no 511"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R1=400 #ohm\n",
+ "R2=800.0\n",
+ "R3=10\n",
+ "V=6\n",
+ "R11=10000.0\n",
+ "R22=400\n",
+ "\n",
+ "#Calculation\n",
+ "Rt=R1+R2+R3\n",
+ "I=V/Rt\n",
+ "Rp=(R11*R22)/(R11+R22)\n",
+ "R=Rp+800\n",
+ "I1=V/R\n",
+ "Vab=I1*Rp\n",
+ "\n",
+ "#Result\n",
+ "print\"Hence the voltmeter will read\", round(Vab,2),\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Hence the voltmeter will read 1.95 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 95
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.41 Page no 512"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "V=2 #V\n",
+ "R=2000.0 #ohm\n",
+ "\n",
+ "#Calculation\n",
+ "I=V/R\n",
+ "pd=I*R\n",
+ "\n",
+ "#Result\n",
+ "print\"Reading of ammeter is\", I*10**3,\"mA \\nReading of voltmeter is\",pd,\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Reading of ammeter is 1.0 mA \n",
+ "Reading of voltmeter is 2.0 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 101
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.42 Page no 512"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "E=3\n",
+ "G=100\n",
+ "R=200.0\n",
+ "n=30\n",
+ "\n",
+ "#Calculation\n",
+ "Ig=E/(G+R)\n",
+ "K=(Ig/n)*10**6\n",
+ "\n",
+ "#Result\n",
+ "print\"Figure of merit of the galvanometer is\", round(K,1),\"micro A/division\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Figure of merit of the galvanometer is 333.3 micro A/division\n"
+ ]
+ }
+ ],
+ "prompt_number": 108
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.43 Page no 513"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "V1=60 #ohm\n",
+ "V2=30\n",
+ "R=300.0\n",
+ "R1=1200\n",
+ "R2=400 #ohm\n",
+ "\n",
+ "#Calculation\n",
+ "V=V1-V2\n",
+ "I=V/R\n",
+ "R11=(R1*R)/(R1+R)\n",
+ "I=V1/(R11+R2)\n",
+ "V11=I*R11\n",
+ "\n",
+ "#Result\n",
+ "print\"Voltmeter will read\", V11,\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Voltmeter will read 22.5 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 115
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.44 Page no 513"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R=20.0 #K ohm\n",
+ "R2=1 #K ohm\n",
+ "\n",
+ "#Calculation\n",
+ "Vr=(R*R2)/(R+R2)\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Voltmeter resistance is\", R,\"K ohm\"\n",
+ "print\"(ii) Voltmeter resistance is\",R2,\"K ohm\"\n",
+ "print\"(iii) Voltmeter resistance is\",round(Vr,2),\"K ohm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Voltmeter resistance is 20.0 K ohm\n",
+ "(ii) Voltmeter resistance is 1 K ohm\n",
+ "(iii) Voltmeter resistance is 0.95 K ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 123
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.45 Page no 514"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "s=20*10**-6\n",
+ "n=30\n",
+ "I=1 #A\n",
+ "G=25 #ohm\n",
+ "\n",
+ "#Calculation\n",
+ "Ig=s*n\n",
+ "S=Ig*G/(1-Ig)\n",
+ "Ra=G*S/(G+S)\n",
+ "\n",
+ "#Result\n",
+ "print\"Resistance of ammeter is\",Ra,\"ohm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Resistance of ammeter is 0.015 ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 128
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
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