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| author | kinitrupti | 2017-05-12 18:40:35 +0530 |
|---|---|---|
| committer | kinitrupti | 2017-05-12 18:40:35 +0530 |
| commit | d36fc3b8f88cc3108ffff6151e376b619b9abb01 (patch) | |
| tree | 9806b0d68a708d2cfc4efc8ae3751423c56b7721 /modern_physics_by_Satish_K._Gupta | |
| parent | 1b1bb67e9ea912be5c8591523c8b328766e3680f (diff) | |
| download | Python-Textbook-Companions-d36fc3b8f88cc3108ffff6151e376b619b9abb01.tar.gz Python-Textbook-Companions-d36fc3b8f88cc3108ffff6151e376b619b9abb01.tar.bz2 Python-Textbook-Companions-d36fc3b8f88cc3108ffff6151e376b619b9abb01.zip | |
Revised list of TBCs
Diffstat (limited to 'modern_physics_by_Satish_K._Gupta')
36 files changed, 0 insertions, 21781 deletions
diff --git a/modern_physics_by_Satish_K._Gupta/chap1.ipynb b/modern_physics_by_Satish_K._Gupta/chap1.ipynb deleted file mode 100755 index 36a74a3e..00000000 --- a/modern_physics_by_Satish_K._Gupta/chap1.ipynb +++ /dev/null @@ -1,462 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:1a5424f1a289fa2ea02065679fbda8bfa43c7eec69070520139086e97a973104"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 1 Coulombs Law"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 1.1 Page no 12"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "q=-3*10**-7 #C\n",
- "e=-1.6*10**-19 #C\n",
- "\n",
- "#Calculation\n",
- "n=q/e\n",
- "\n",
- "#Result\n",
- "print\"Number of electrons transferred from wool to polythene is\", n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Number of electrons transferred from wool to polythene is 1.875e+12\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 1.2 Page no 12"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "m=3.11 #g\n",
- "Z=29\n",
- "A=63.5 \n",
- "N=6.023*10**23\n",
- "e=1.6*10**-19\n",
- "\n",
- "#Calculation\n",
- "n=(N*m)/A\n",
- "n1=n*Z\n",
- "q=n1*e\n",
- "\n",
- "#Result\n",
- "print\"Total positive or negative charge is\", round(q*10**-5,2),\"*10**5 C\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Total positive or negative charge is 1.37 *10**5 C\n"
- ]
- }
- ],
- "prompt_number": 23
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 1.3 Page no 12"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "q1=2*10**-7\n",
- "q2=3*10**-7\n",
- "r=0.3 #m\n",
- "a=9*10**9\n",
- "\n",
- "#Calculation\n",
- "F=(a*q1*q2)/r**2\n",
- "\n",
- "#Result\n",
- "print\"Force between two small charged spheres is\", F*10**3,\"*10**-3\",\"N(repulsive\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Force between two small charged spheres is 6.0 *10**-3 N(repulsive\n"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 1.4 page no. 12"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "F=3.7*10**-9 #N\n",
- "r=5*10**-10 #m\n",
- "a=9*10**9\n",
- "q1=1.6*10**-19\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "n=math.sqrt(F*r**2/(a*q1**2))\n",
- "\n",
- "#Result\n",
- "print\"number of electrons is\", round(n,0)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "number of electrons is 2.0\n"
- ]
- }
- ],
- "prompt_number": 16
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 1.5 Page no 12"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "q1=0.4*10**-6 #C\n",
- "q2=0.8*10**-6 #C\n",
- "F12=0.2 #N\n",
- "a=9.0*10**9\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "r=math.sqrt((a*q1*q2)/F12)\n",
- "\n",
- "#Result\n",
- "print\"(a) Distance between two spheres is\", r,\"m\"\n",
- "print\"(b) Force on charge q2 due to q1 is\",F12,\"N\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a) Distance between two spheres is 0.12 m\n",
- "(b) Force on charge q2 due to q1 is 0.2 N\n"
- ]
- }
- ],
- "prompt_number": 32
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 1.6 Page no 13"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "q1=5*10**-8 #C\n",
- "m1=8*10**-3 #Kg\n",
- "a=9*10**9\n",
- "r=0.05 #m\n",
- "\n",
- "#Calculation\n",
- "q2=m1*9.8*r**2/(a*q1)\n",
- "\n",
- "#Result\n",
- "print\"Charge q2 is\", round(q2*10**7,2)*10**-7,\"C(positive)\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Charge q2 is 4.36e-07 C(positive)\n"
- ]
- }
- ],
- "prompt_number": 40
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 1.7 Page no 13"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "q1=6.5*10**-7 #C\n",
- "q2=6.5*10**-7\n",
- "r=0.5 #m\n",
- "a=9*10**9\n",
- "K=80.0\n",
- "\n",
- "#Calculation\n",
- "Fair=a*q1*q2/r**2\n",
- "r1=0.5/2.0\n",
- "F1=a*4*q1*q2/r1**2\n",
- "Fwater=Fair/K\n",
- "\n",
- "#Result\n",
- "print\"(a) Mutual force of electrostatic repulsion is\", Fair*10**2,\"*10**-2 N\"\n",
- "print\"(b) (i) Force of repulsion is\", round(F1,4),\"N\"\n",
- "print \"(ii) Force of repulsion is\",round(Fwater*10**4,1),\"*10**-4 N\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a) Mutual force of electrostatic repulsion is 1.521 *10**-2 N\n",
- "(b) (i) Force of repulsion is 0.2434 N\n",
- "(ii) Force of repulsion is 1.9 *10**-4 N\n"
- ]
- }
- ],
- "prompt_number": 57
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 1.8 Page no 13"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "q1=6.5*10**-7 #C\n",
- "r=0.05 #m\n",
- "a=9*10**9\n",
- "r1=0.5\n",
- "\n",
- "#Calculation\n",
- "q11=q1/2.0\n",
- "q21=(q1+q11)/2.0\n",
- "F=(a*q11*q21)/r1**2\n",
- "\n",
- "#Result\n",
- "print\"New force of repulsion between A and B is\", round(F*10**3,3),\"*10**-3 N\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "New force of repulsion between A and B is 5.704 *10**-3 N\n"
- ]
- }
- ],
- "prompt_number": 64
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 1.10 Page no 14"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "a=9.0*10**9\n",
- "r=0.2\n",
- "m=9.8*10**-3\n",
- "a1=0.1\n",
- "a2=0.5\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "a11=m*(a1/(math.sqrt(a2**2-a1**2)))\n",
- "q=math.sqrt((a11*r**2)/a)\n",
- "\n",
- "#Result\n",
- "print\"Charge on each ball is\", round(q*10**8,2)*10**-6,\"C\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Charge on each ball is 9.43e-06 C\n"
- ]
- }
- ],
- "prompt_number": 100
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 1.12 Page no 14"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "qa=10**-5 #C\n",
- "qb=5*10**-6 #C\n",
- "qc=-5*10**-6 #C\n",
- "r=0.1 #m\n",
- "a=9*10**9\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "Fab=(a*qa*qb)/r**2\n",
- "Fac=Fab\n",
- "F=math.sqrt(Fab**2+Fac**2+(2*Fab*Fac*math.cos(120*3.14/180.0)))\n",
- "\n",
- "#Result\n",
- "print\"Resultant force is\", round(F,0),\"N\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Resultant force is 45.0 N\n"
- ]
- }
- ],
- "prompt_number": 75
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 1.13 Page no 15"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "qa=1\n",
- "qb=100\n",
- "ab=10\n",
- "a=9*10**9\n",
- "qd=75\n",
- "a1=5\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "Fab=(a*qa*qb)/ab**2\n",
- "Fac=Fab\n",
- "Fac1=(a*qa*qd)/(ab**2-a1**2)\n",
- "Fx=Fab*math.cos(60*3.14/180.0)+Fac1*math.cos(60*3.14/180.0)\n",
- "Fy=Fac\n",
- "F=math.sqrt(Fx**2+Fy**2)\n",
- "B=Fy/Fx\n",
- "B1=math.atan(B)*180/3.14\n",
- "\n",
- "#Result\n",
- "print\"Resultant force on charge qa is inclined at\", round(B1,0),\"Degree\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Resultant force on charge qa is inclined at 45.0 Degree\n"
- ]
- }
- ],
- "prompt_number": 90
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/modern_physics_by_Satish_K._Gupta/chap10.ipynb b/modern_physics_by_Satish_K._Gupta/chap10.ipynb deleted file mode 100755 index 1207ccc1..00000000 --- a/modern_physics_by_Satish_K._Gupta/chap10.ipynb +++ /dev/null @@ -1,240 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:59cd979321ac17257305319649b46031fe00ee56620ef7034e9b0708b633af30"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 10 Thermoelectric effect"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 10.1 Page no 304"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "a=15 #degree C\n",
- "b=270 \n",
- "\n",
- "#Calculation\n",
- "c=(2*b)-a\n",
- "\n",
- "#Result\n",
- "print\"Temperature is\", c,\"degree C\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Temperature is 525 degree C\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 10.2 Page no 304"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "a=10 #degree C\n",
- "a1=527 \n",
- "\n",
- "#Calculation\n",
- "a2=(a+a1)/2.0\n",
- "print\"Temperature is\",a2,\"degree C\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Temperature is 268.5 degree C\n"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 10.3 Page no 304"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "R=100 #ohm\n",
- "I=10**-6 #A\n",
- "E=40.0*10**-6 #V\n",
- "\n",
- "#Calculaton\n",
- "e=R*I\n",
- "t=e/E\n",
- "\n",
- "#Result\n",
- "print\"Smallest temperature difference is\",t,\"degree C\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Smallest temperature difference is 2.5 degree C\n"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 10.4 Page no 304"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "a=80 #Degree C\n",
- "b=13.31 #micro V degree/C\n",
- "B=-0.038 #micro V degree/C**2\n",
- "\n",
- "#Calculation\n",
- "E=(b*a)+((B*a**2)/2.0)\n",
- "an=-b/B\n",
- "\n",
- "#Result\n",
- "print\"Thermo emf of silver is\",E,\"micro V\"\n",
- "print\"Neutral temperature is\",round(an,2),\"degree C\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Thermo emf of silver is 943.2 micro V\n",
- "Neutarl temperature is 350.26 degree C\n"
- ]
- }
- ],
- "prompt_number": 13
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 10.5 Page no 304"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "a=14 #micro V degree/C\n",
- "b=-0.04 #micro V degree/C**2\n",
- "\n",
- "#Calculation\n",
- "an=-a/b\n",
- "\n",
- "#Result\n",
- "print\"Neutral temperature is\",an,\"degree C\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Neutral temperature is 350.0 degree C\n"
- ]
- }
- ],
- "prompt_number": 16
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 10.6 Page no 304"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "E=500 #micro V\n",
- "a=41 #micro V degree/C\n",
- "b=0.041 #micro V degree/C**2\n",
- "c=-5500\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "A=-(a+(math.sqrt(a**2-4*b*c)))/(2.0*b)\n",
- "\n",
- "#Result\n",
- "print\"Temperature of hot junction is\", round(A,2),\"degree C\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Temperature of hot junction is -1119.8 degree C\n"
- ]
- }
- ],
- "prompt_number": 24
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/modern_physics_by_Satish_K._Gupta/chap11.ipynb b/modern_physics_by_Satish_K._Gupta/chap11.ipynb deleted file mode 100755 index bce9c399..00000000 --- a/modern_physics_by_Satish_K._Gupta/chap11.ipynb +++ /dev/null @@ -1,466 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:516acdbb9c17332f6e1ea973b0354e9af84e691b9ea2bf4ff9f153c48f9c31a8"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 11 Magnetic Effects Of Current And Magnetism "
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 11.1 Page no 350"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given \n",
- "i=90 #A\n",
- "a=1.5 #m\n",
- "b=2\n",
- "l=10**7\n",
- "\n",
- "#Calculation\n",
- "u=l*((b*i)/a)\n",
- "\n",
- "#Result\n",
- "print\"Magnitude of the magnetic field is\",u*10**-9,\"10**-5\",\"T\"\n",
- "print\"Direction of the magnetic field is south\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Magnitude of the magnetic field is 1.2 10**-5 T\n",
- "Direction of the magnetic field is south\n"
- ]
- }
- ],
- "prompt_number": 10
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 11.2 Page no 350"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "I=3 #A\n",
- "a=0.15 #m\n",
- "e=10**-7\n",
- "b=2\n",
- "\n",
- "#Calculation\n",
- "Z=(e*b*I)/a\n",
- "\n",
- "#Result\n",
- "print\"The magnitude of the magnetic field is\",Z*10**6,\"10**-6\",\"T\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The magnitude of the magnetic field is 4.0 10**-6 T\n"
- ]
- }
- ],
- "prompt_number": 16
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 11.4 Page no 351"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "l=10 #A\n",
- "a=0.05 #cm\n",
- "b=10**-7\n",
- "a1=45 #degree\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "B=b*(l/a)*(math.sin(a1*3.14/180.0)+math.sin(a1*3.14/180.0))\n",
- "B1=4*B\n",
- "\n",
- "#Result\n",
- "print\"Magnetic field induction is\",round(B1*10**4,2),\"*10**-4 T\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Magnetic field induction is 1.13 *10**-4 T\n"
- ]
- }
- ],
- "prompt_number": 38
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 11.5 Page no 351"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "n=10\n",
- "a=0.08 #m\n",
- "l=2 #A\n",
- "v=10**-7\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "B=(v*l*math.pi*n*l)/a\n",
- "\n",
- "#Result\n",
- "print\"Magnitude of the magnetic field is\", round(B*10**4,2),\"*10**-4 T\"\n",
- "print\"Direction of the field is vertically downward.\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Magnitude of the magnetic field is 1.57 *10**-4 T\n",
- "Direction of the field is vertically downward.\n"
- ]
- }
- ],
- "prompt_number": 44
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 11.6 Page no 351"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "v=2.2*10**6 #m/s\n",
- "a=0.5*10**-10 #m\n",
- "e=1.6*10**-19\n",
- "m=10**-7 #N/a**2\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "I=e*v/(2*math.pi*a)\n",
- "B=m*2*math.pi*I/a\n",
- "\n",
- "#Result\n",
- "print\"Magnetic field produced at the centre is\", B,\"T\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Magnetic field produced at the centre is 14.08 T\n"
- ]
- }
- ],
- "prompt_number": 49
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 11.8 Page no 352"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "I=10 #A\n",
- "a=0.03 #m\n",
- "a1=360 #degree\n",
- "a2=90\n",
- "a3=10**-7\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "A=a1-a2\n",
- "B=a3*I*3*math.pi/(a*2.0)\n",
- "\n",
- "#Result\n",
- "print\"Magnitude of magnetic field is\", round(B*10**4,2),\"*10**-4 T\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Magnitude of magnetic field is 1.57 *10**-4 T\n"
- ]
- }
- ],
- "prompt_number": 55
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 11.10 Page no 352"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "n=500\n",
- "I=7 #A\n",
- "a=0.05 #m\n",
- "x=0.12 #m\n",
- "m=10**-7\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "B=(2*math.pi*n*I*a**2)/(a**2+x**2)**1.5\n",
- "B1=m*2*math.pi*n*I/a\n",
- "\n",
- "#Result\n",
- "print\"(a) Magnetic field at the point on the axis is\", round(B*10**-4,1),\"*10**-3 tesla\"\n",
- "print\"(b)Magnetic field at the centre of the coil is\",round(B1*10**2,2),\"*10**-2 tesla\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a) Magnetic field at the point on the axis is 2.5 *10**-3 tesla\n",
- "(b)Magnetic field at the centre of the coil is 4.4 *10**-2 tesla\n"
- ]
- }
- ],
- "prompt_number": 69
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 11.13 Page no 353"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "B=2.52*10**-3 #T\n",
- "l=0.5 #m\n",
- "N=500\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "n=N/l\n",
- "I=B/(4*math.pi*10**-7*n)\n",
- "\n",
- "#Result\n",
- "print\"Current in the solenoid is\", round(I,0),\"A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Current in the solenoid is 2.0 A\n"
- ]
- }
- ],
- "prompt_number": 75
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 11.14 Page no 353"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "I=5 #A\n",
- "l=2.0 #m\n",
- "n=1000\n",
- "n1=5\n",
- "\n",
- "#Calculation\n",
- "N=n*n1\n",
- "n2=N/l\n",
- "B=4*math.pi*10**-7*n2*I\n",
- "\n",
- "#Result\n",
- "print\"Magnetic field at the center is\", round(B*10**2,2),\"*10**-2 T\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Magnetic field at the center is 1.57 *10**-2 T\n"
- ]
- }
- ],
- "prompt_number": 80
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 11.15 Page no 353"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "a=16.0*10**-2 #m\n",
- "n=20\n",
- "I=16 #A\n",
- "b=10**-7\n",
- "a1=10.0*10**-2\n",
- "n1=25\n",
- "I1=18\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "B=b*2*math.pi*n*I/a\n",
- "B1=b*2*math.pi*n1*I1/a1\n",
- "B2=B1-B\n",
- "\n",
- "#Result\n",
- "print\"Magnitude of the magnetic field is\", round(B1*10**4,2),\"*10**-4 T\"\n",
- "print\"Direction is towards west\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Magnitude of the magnetic field is 28.27 *10**-4 T\n",
- "Direction is towards west\n"
- ]
- }
- ],
- "prompt_number": 88
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 11.18 Page no 355"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "l=11 #A\n",
- "s=3500\n",
- "r1=25 #cm\n",
- "r2=26\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "w=((r1+r2)/2.0)\n",
- "l1=2*math.pi*w\n",
- "n=s/l1\n",
- "B=4*math.pi*10**-7*n*l\n",
- "\n",
- "#Result\n",
- "print\"(a) Magnetic field outside the toroid is zero\"\n",
- "print\"(b) Magnetic field inside the core is\",round(B*10**4,2),\"*10**-2 T\"\n",
- "print\"(c) Magnetic field in the empty space is zero\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a) Magnetic field outside the toroid is zero\n",
- "(b) Magnetic field inside the core is 3.02 *10**-2 T\n",
- "(c) Magnetic field in the empty space is zero\n"
- ]
- }
- ],
- "prompt_number": 105
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/modern_physics_by_Satish_K._Gupta/chap12.ipynb b/modern_physics_by_Satish_K._Gupta/chap12.ipynb deleted file mode 100755 index 30e711a0..00000000 --- a/modern_physics_by_Satish_K._Gupta/chap12.ipynb +++ /dev/null @@ -1,1435 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:48ced6bcb2294321844f4027bedbd6f874bf0da0aa33534190457acb1015aba8"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 12 Motion of charged particle"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.1 Page no 379"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "m=1.6*10**-27 #Kg\n",
- "e=1.6*10**-19\n",
- "Ey=2*10**4 #V/m\n",
- "x=0.1 #m\n",
- "vx=5*10**6 #m/s\n",
- "\n",
- "#Calculation\n",
- "t=x/vx\n",
- "Fy=e*Ey\n",
- "a=Fy/m\n",
- "y=a*t**2/2.0\n",
- "\n",
- "#Result\n",
- "print\"Transverse deflection is\", y*10**3,\"mm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Transverse deflection is 0.4 mm\n"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.2 Page no 379"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "n=8*10**28 #/m**2\n",
- "l=1 #m\n",
- "A=8*10**-6 #m**2\n",
- "B=5*10**-3 #T\n",
- "F=8*10**-2 #N\n",
- "e=1.6*10**-19 #C\n",
- "\n",
- "#Calculation\n",
- "vd=F/(B*n*A*l*e)\n",
- "\n",
- "#Result\n",
- "print\"Drift velocity is\", vd*10**4,\"*10**-4 m/s\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Drift velocity is 1.5625 *10**-4 m/s\n"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.3 Page no 379"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "q=1.6*10**-19 #C\n",
- "v=10**7 #m/s\n",
- "B=3 #T\n",
- "\n",
- "#Calculation\n",
- "F=q*v*B\n",
- "\n",
- "#Result\n",
- "print\"Instantaneous force is\", F,\"N\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Instantaneous force is 4.8e-12 N\n"
- ]
- }
- ],
- "prompt_number": 14
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.4 Page no 379"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "I=2 #A\n",
- "a=0.1 #m\n",
- "u=10**-7\n",
- "q=1.6*10**-19\n",
- "v=4*10**4 #m/s\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "B=(u*2*I)/a\n",
- "F=B*q*v\n",
- "\n",
- "#Result\n",
- "print\"Force of magnetic field is\", F,\"N\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Force of magnetic field is 2.56e-20 N\n"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.6 Page no 380"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "v=10**5 #m/s\n",
- "e=1.6*10**-19 #C\n",
- "m=9.1*10**-31 #Kg\n",
- "B=0.019*10**-4 #T\n",
- "\n",
- "#Calculation\n",
- "r=m*v/(B*e)\n",
- "\n",
- "#Result\n",
- "print\"Radius of the circular path is\", round(r,3),\"m\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Radius of the circular path is 0.299 m\n"
- ]
- }
- ],
- "prompt_number": 10
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.7 Page no 380"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "e=1.6*10**-19\n",
- "m=9*10**-31 #Kg\n",
- "T=10**-6 #S\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "B=2*math.pi*m/(e*T)\n",
- "\n",
- "#Result\n",
- "print\"Magnetic field is\", round(B*10**5,3)*10**-5 ,\"T\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Magnetic field is 3.534e-05 T\n"
- ]
- }
- ],
- "prompt_number": 16
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.9 Page no 381"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "m=1.67*10**-27 #Kg\n",
- "e=1.60*10**-19\n",
- "V=10**7 #Hz\n",
- "R=0.6 #m\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "B=2*math.pi*m*V/e\n",
- "Emax=(B**2*e**2*R**2/(2*m))/1.6*10**13\n",
- "\n",
- "#Result\n",
- "print\"Kinetic energy of the proton is\",round(Emax,3),\"Mev\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Kinetic energy of the proton is 7.417 Mev\n"
- ]
- }
- ],
- "prompt_number": 25
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.10 Page no 381"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "I=5 #A\n",
- "l=0.1 #m\n",
- "m=3*10**-3\n",
- "g=9.8\n",
- "a=0.5\n",
- "\n",
- "#Calculation\n",
- "w=m*g*l\n",
- "B=w/a\n",
- "\n",
- "#Result\n",
- "print\"Magnitude of the magnetic field is\", B*10**3,\"*10**-3 tesla\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Magnitude of the magnetic field is 5.88 *10**-3 tesla\n"
- ]
- }
- ],
- "prompt_number": 29
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.11 Page no 381"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#given\n",
- "I1=4 #A\n",
- "I2=6\n",
- "r=0.03 #m\n",
- "u=10**-7\n",
- "\n",
- "#Calculation\n",
- "F=u*2*I1*I2/r\n",
- "\n",
- "#Result\n",
- "print\"Force per unit length is\", F*10**4,\"*10**-4 N/m\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Force per unit length is 1.6 *10**-4 N/m\n"
- ]
- }
- ],
- "prompt_number": 33
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.12 Page no 381"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "I1=5 #A\n",
- "I2=12\n",
- "r=0.4 #m\n",
- "u=10**-7\n",
- "\n",
- "#Calculation\n",
- "F=u*2*I1*I2/r\n",
- "F1=u*2*I1*I2/r\n",
- "\n",
- "#Result\n",
- "print\"(i) Force when current flows in same direction is\", F,\"N/m\"\n",
- "print\"(ii) Force when current flows in opposite direction is\",F1,\"N/m\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(i) Force when current flows in same direction is 3e-05 N/m\n",
- "(ii) Force when current flows in opposite direction is 3e-05 N/m\n"
- ]
- }
- ],
- "prompt_number": 38
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.13 Page no 381"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "I1=300 #A\n",
- "r=1.5*10**-2 #m\n",
- "u=10**-7\n",
- "\n",
- "#Calculation\n",
- "F=u*2*I1*I1/r\n",
- "\n",
- "#Result\n",
- "print\"Force per unit length is\",F,\"N/m\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Force per unit length is 1.2 N/m\n"
- ]
- }
- ],
- "prompt_number": 40
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.14 Page no 381"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "I=10 #A\n",
- "n=100\n",
- "A=8*10**-2 #m**2\n",
- "B=5 #T\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "t=n*B*I*A*math.cos(60*3.14/180.0)\n",
- "\n",
- "#Result\n",
- "print\"Torque is\", round(t,0),\"Nm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Torque is 200.0 Nm\n"
- ]
- }
- ],
- "prompt_number": 44
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.15 Page no 381"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "n=30\n",
- "I=6 #A\n",
- "B=1 #T\n",
- "r=8*10**-2 #m\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "A=math.pi*r**2\n",
- "t=n*B*I*A*math.sin(60*3.14/180.0)\n",
- "\n",
- "#Result\n",
- "print\"(a) Magnitude of the counter torque is\", round(t,3),\"Nm\"\n",
- "print\"(b) Torque on the planar loop is independent of its shape, the torque will remain unchanged.\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a) Magnitude of the counter torque is 3.133 Nm\n",
- "(b) Torque on the planar loop is independent of its shape, the torque will remain unchanged.\n"
- ]
- }
- ],
- "prompt_number": 50
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.16 Page no 382"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "B=100*10**-4 #T\n",
- "I=10 #A\n",
- "l=44\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "r=l/(2.0*math.pi)\n",
- "A=math.pi*r**2\n",
- "t=B*I*A\n",
- "\n",
- "#Result\n",
- "print\"Maximum torque is\", round(t*10**-1,2),\"*10**-3 Nm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Maximum torque is 1.54 *10**-3 Nm\n"
- ]
- }
- ],
- "prompt_number": 62
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.17 Page no 382"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "n=20\n",
- "r=10*10**-2 #m\n",
- "B=0.10 #T\n",
- "I=5 #A\n",
- "n1=10**29 #/m**3\n",
- "A1=10**-5 #m**2\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "A=math.pi*r**2\n",
- "t=n*B*I*A*math.sin(0*3.14/180.0)\n",
- "F=B*I/(n1*A1)\n",
- "\n",
- "#Result\n",
- "print\"(a) Total torque on the coil is\", t\n",
- "print\"(b) Net force on a planar loop in a magnetic field is always zero\"\n",
- "print\"(c) Average force is\",F,\"N\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a) Total torque on the coil is 0.0\n",
- "(b) Net force on a planar loop in a magnetic field is always zero\n",
- "(c) Average force is 5e-25 N\n"
- ]
- }
- ],
- "prompt_number": 68
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.18 Page no 382"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "A=5*10**-4 #m**2\n",
- "n=60\n",
- "a=18 #degree\n",
- "B=90*10**-4 #T\n",
- "I=0.20*10**-3 #A\n",
- "\n",
- "#calculation\n",
- "k=n*B*I*A/a\n",
- "\n",
- "#Result\n",
- "print\"Torsional constant is\",k,\"N m per degree\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Torsional constant is 3e-09 N m per degree\n"
- ]
- }
- ],
- "prompt_number": 70
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.21 Page no 383"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "G=15 #ohm\n",
- "Ig=2*10**-3 #A\n",
- "I=5 #A\n",
- "\n",
- "#Calculation\n",
- "S=Ig*G/(I-Ig)\n",
- "\n",
- "#Result\n",
- "print\"Shunt resistance is\",round(S,3),\"Ohm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Shunt resistance is 0.006 Ohm\n"
- ]
- }
- ],
- "prompt_number": 74
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.22 Page no 383"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "V=50*10**-3 #V\n",
- "G=100.0 #ohm\n",
- "\n",
- "#Calculation\n",
- "Ig=V/G\n",
- "S=Ig*G/(I-Ig)\n",
- "\n",
- "#Result\n",
- "print\"Shunt resistance is\", round(S,2),\"ohm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Shunt resistance is 0.01 ohm\n"
- ]
- }
- ],
- "prompt_number": 79
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.23 Page no 383"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "G=100 #ohm\n",
- "Ig=5*10**-3 #A\n",
- "I=5 #A\n",
- "\n",
- "#Calculation\n",
- "S=Ig*G/(I-Ig)\n",
- "\n",
- "#Result\n",
- "print\"Shunt resistance is\",S"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Shunt resistance is 0.1001001001\n"
- ]
- }
- ],
- "prompt_number": 80
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.24 Page no 383"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "G=5 #ohm\n",
- "Ig=15.0*10**-3 #A\n",
- "I=1.5\n",
- "V=1.5 #V\n",
- "\n",
- "#Calculation\n",
- "S=Ig*G/(I-Ig)\n",
- "R=(V/Ig)-G\n",
- "\n",
- "#Result\n",
- "print\"(a) To enable galvanometer to read 1.5 A is\", round(S,2),\"ohm\"\n",
- "print\"(b) To enable galvanometer to read 1.5 V is\",R,\"ohm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a) To enable galvanometer to read 1.5 A is 0.05 ohm\n",
- "(b) To enable galvanometer to read 1.5 V is 95.0 ohm\n"
- ]
- }
- ],
- "prompt_number": 88
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.25 Page no 383"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "G=10 #ohm\n",
- "Ig=25.0*10**-3 #A\n",
- "V=120 #V\n",
- "I=20 #A\n",
- "\n",
- "#Calculation\n",
- "R=(V/Ig)-G\n",
- "S=Ig*G/(I-Ig)\n",
- "\n",
- "#Result\n",
- "print\"(a) To convert the galvanometer into the voltmeter reading is\" ,R,\"ohm\"\n",
- "print\"(b) To convert the galvanometer into the ammeter reading is\",round(S,4),\"ohm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a) To convert the galvanometer into the voltmeter reading is 4790.0 ohm\n",
- "(b) To convert the galvanometer into the ammeter reading is 0.0125 ohm\n"
- ]
- }
- ],
- "prompt_number": 95
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.26 Page no 383"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "E=3 #v\n",
- "R=55 #ohm\n",
- "Ra=1\n",
- "I=50*10**-3 #A\n",
- "\n",
- "#Calculation\n",
- "r=(E/I)-(R+Ra)\n",
- "\n",
- "#Result\n",
- "print\"Value of r is\", r,\"ohm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Value of r is 4.0 ohm\n"
- ]
- }
- ],
- "prompt_number": 98
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.27 Page no 384"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "E=60 #V\n",
- "R1=400 #ohm\n",
- "R2=300\n",
- "V1=30.0\n",
- "a=120000\n",
- "\n",
- "#Calculation\n",
- "Rv=(-V1*a)/(V1*(R1+R2)-E*R1)\n",
- "R=Rv*R2/(Rv+R2)\n",
- "I1=E/(R+R1)\n",
- "V=I1*R\n",
- "\n",
- "#Result\n",
- "print\"Voltmeter reads\", V,\"V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Voltmeter reads 22.5 V\n"
- ]
- }
- ],
- "prompt_number": 108
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.28 Page no 384"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "Rv=400.0\n",
- "E=84 #V\n",
- "r=100.0\n",
- "r1=200\n",
- "\n",
- "#Calculation\n",
- "R=1/(1/Rv+1/100.0)\n",
- "R1=R+200\n",
- "I=E/R1\n",
- "I1=I/5.0\n",
- "V=I1*Rv\n",
- "R2=r+r1\n",
- "I2=E/R2\n",
- "V2=I2*r\n",
- "\n",
- "#Result\n",
- "print\"(a) Reading on voltmeter is\", V,\"V\"\n",
- "print\"(b) Potential difference is\",V2,\"V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a) Reading on voltmeter is 24.0 V\n",
- "(b) Potential difference is 28.0 V\n"
- ]
- }
- ],
- "prompt_number": 122
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.29 Page no 385"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "E=120 #v\n",
- "Rv=10**4 #ohm\n",
- "a=4.0\n",
- "\n",
- "#Calculation\n",
- "x=(Rv*(E-a))/a\n",
- "\n",
- "#Result\n",
- "print\"Large resistance is\", x*10**-3,\"K ohm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Large resistance is 290.0 K ohm\n"
- ]
- }
- ],
- "prompt_number": 131
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.30 Page no 385"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "B=0.75 #T\n",
- "E=9*10**5 #V/m\n",
- "V=15*10**3\n",
- "\n",
- "#Calculation\n",
- "v=E/B\n",
- "a=v**2/(2.0*V)\n",
- "\n",
- "#Result\n",
- "print\"The value of e/m is\", a*10**-7,\"*10**7 C/Kg\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of e/m is 4.8 *10**7 C/Kg\n"
- ]
- }
- ],
- "prompt_number": 127
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.31 Page no 385"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "m=9.11*10**-31 #Kg\n",
- "e=1.60*10**-19 #C\n",
- "B=0.40*10**-4 #T\n",
- "a=18*1.6*10**-16\n",
- "PQ=0.30\n",
- "a2=1.52 #degree\n",
- "\n",
- "#calculation\n",
- "import math\n",
- "r=math.sqrt(2*m*a)/(B*e)\n",
- "a1=(PQ/r)\n",
- "PA=r*(1-math.cos(a2*3.14/180.0))\n",
- "\n",
- "#Result\n",
- "print\"up and down deflection of the beam is\", round(PA*10**3,0),\"mm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "up and down deflection of the beam is 4.0 mm\n"
- ]
- }
- ],
- "prompt_number": 153
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.32 Page no 386"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "m=60*10**-3\n",
- "g=9.8\n",
- "I=5\n",
- "l=0.45\n",
- "\n",
- "#Calculation\n",
- "B=m*g/(I*l)\n",
- "T=2*m*g\n",
- "\n",
- "#Result\n",
- "print\"(i) Magnetic field is\", round(B,3),\"T\"\n",
- "print\"(ii) Total tension in the wire is\",T,\"N\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(i) Magnetic field is 0.261 T\n",
- "(ii) Total tension in the wire is 1.176 N\n"
- ]
- }
- ],
- "prompt_number": 161
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.33 Page no 386"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "B=0.15 #T\n",
- "m=0.30 #Kg/m\n",
- "a=30 #degree\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": 165
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.34 Page no 386"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "I1=4 #A\n",
- "I2=3\n",
- "r=3.0*10**-2 #m\n",
- "u=10**-7\n",
- "l=5*10**-2\n",
- "\n",
- "#Calculation\n",
- "F=u*2*I1*I2/r\n",
- "F1=F*l\n",
- "\n",
- "#Result\n",
- "print\"Total force is\", F1,\"N (attractive force)\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Total force is 4e-06 N (attractive force)\n"
- ]
- }
- ],
- "prompt_number": 171
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.35 Page no 386"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "AB=25*10**-2 #m\n",
- "BC=10*10**-2\n",
- "r1=2.0*10**-2 #m\n",
- "I1=15 #A\n",
- "I2=25 #A\n",
- "u=10**-7\n",
- "\n",
- "#Calculation\n",
- "r2=BC+r1\n",
- "F1=u*2*I1*I2*AB/r1\n",
- "F2=u*2*I1*I2*AB/r2\n",
- "F=F1-F2\n",
- "\n",
- "#Result\n",
- "print\"Net force on the loop is\", F*10**4,\"*10**-4 N (towards XY)\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Net force on the loop is 7.8125 *10**-4 N (towards XY)\n"
- ]
- }
- ],
- "prompt_number": 182
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.36 Page no 387"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "M=30*10**-3 #Kg\n",
- "g=9.8 #m/s**2\n",
- "l=0.5 #m\n",
- "r=10**-2\n",
- "u=10**-7\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "I=math.sqrt((M*g*r)/(u*2*l))\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 171.46 A\n"
- ]
- }
- ],
- "prompt_number": 186
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.37 Page no 387"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "n=900\n",
- "l=0.6\n",
- "u=10**-7\n",
- "l2=0.02 #m\n",
- "l1=6\n",
- "m=2.5*10**-3 #Kg\n",
- "g=9.8\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "n1=n/l\n",
- "B=4*math.pi*n1\n",
- "F=B*l1*l2\n",
- "I=m*g/F\n",
- "\n",
- "#Result\n",
- "print\"Current in the winding of the secondary is\", round(I*10**7,1),\"A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Current in the winding of the secondary is 108.3 A\n"
- ]
- }
- ],
- "prompt_number": 197
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.38 Page no 387"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "A=1.6*10**-3 #m**2\n",
- "n=200\n",
- "B=0.2 #T\n",
- "a=30 #degree\n",
- "K=10**-6 #N m /degree\n",
- "a1=0.1\n",
- "\n",
- "#Calculation\n",
- "Imax=K*a/(n*B*A)\n",
- "Imin=K*a1/(n*B*A)\n",
- "\n",
- "#Result\n",
- "print\"(i) Minimum current is\",round(Imax*10**4,2),\"*10**-4 A\"\n",
- "print\"(ii) Smallest current that can be detected is\",Imin,\"A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(i) Minimum current is 4.69 *10**-4 A\n",
- "(ii) Smallest current that can be detected is 1.5625e-06 A\n"
- ]
- }
- ],
- "prompt_number": 204
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.39 Page no 388 "
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "R=5000.0 #ohm/V\n",
- "V=5\n",
- "V1=20.0\n",
- "\n",
- "#Calculation\n",
- "Ig=1/R\n",
- "G=V/Ig\n",
- "R1=(V1/Ig)-G\n",
- "Rn=R1+G\n",
- "Rv=Rn/V1\n",
- "\n",
- "#Result\n",
- "print\"New voltmeter will be still graded as\", Rv,\"ohm/V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "New voltmeter will be still graded as 5000.0 ohm/V\n"
- ]
- }
- ],
- "prompt_number": 212
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.40 Page no 388"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "V1=100 #V\n",
- "Rv=400.0 \n",
- "\n",
- "#Calculation\n",
- "I1=V1/Rv\n",
- "V=I1*20\n",
- "V2=V1+V\n",
- "V3=V2-V1\n",
- "\n",
- "#Result\n",
- "print\"Error in the reading of the voltmeter is\",V3,\"V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Error in the reading of the voltmeter is 5.0 V\n"
- ]
- }
- ],
- "prompt_number": 215
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/modern_physics_by_Satish_K._Gupta/chap13.ipynb b/modern_physics_by_Satish_K._Gupta/chap13.ipynb deleted file mode 100755 index 13ac3401..00000000 --- a/modern_physics_by_Satish_K._Gupta/chap13.ipynb +++ /dev/null @@ -1,547 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:a82b820d4c986803883c47a06b19e19453881487c8189443b60dab54c2680966"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 13 Magnetic dipole"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 13.1 Page no 415"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "F=14.4*10**-4 #N\n",
- "r=0.05 #m\n",
- "u=10.0**-7\n",
- "F1=1.6*10**-4 #N\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "m=math.sqrt(F*r**2/u)\n",
- "r1=math.sqrt(u*6*6/F1)\n",
- "\n",
- "#Result\n",
- "print\"Distance between the poles is\", r1,\"m\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Distance between the poles is 0.15 m\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 13.2 Page no 415"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "F=0.8*10**-3\n",
- "r=0.1 #m\n",
- "g=9.8\n",
- "u=10.0**-7\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "m=math.sqrt(F*g*r**2/(u*5.0))\n",
- "a=5*m\n",
- "\n",
- "#Result\n",
- "print\"Strength of each pole is\", round(a,1),\"A m\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Strength of each pole is 62.6 A m\n"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 13.4 Page no 416"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "m=50 #Am\n",
- "r=0.2 #m\n",
- "l=0.05 #M\n",
- "M=5 #Am**2\n",
- "u=10**-7\n",
- "\n",
- "#Calculation\n",
- "Baxial=u*2*M*r/((r**2-l**2)**2)\n",
- "Bequi=u*M/((r**2+l**2)**1.5)\n",
- "\n",
- "#Result\n",
- "print\"(i) Magnetic field at a distance of 0.2 m from its centre on axial line is\",round(Baxial*10**4,2),\"*10**-4 T\"\n",
- "print\"(ii) Magnetic field at a distance of 0.2 m from its centre on its equitorial line is\",round(Bequi*10**5,3)*10**-5,\"T\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(i) Magnetic field at a distance of 0.2 m from its centre on axial line is 1.42 *10**-4 T\n",
- "(ii) Magnetic field at a distance of 0.2 m from its centre on its equitorial line is 5.707e-05 T\n"
- ]
- }
- ],
- "prompt_number": 22
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 13.5 Page no 416"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "M=12.5 #Am**2\n",
- "OP=0.05 #m\n",
- "u=10**-7\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "b1=(u*2*M)/(OP**3)\n",
- "b2=(u*M)/(OP**3)\n",
- "B=math.sqrt(b1**2+b2**2)\n",
- "a=b2/b1\n",
- "a1=math.atan(a*180/3.14)\n",
- "\n",
- "#Result\n",
- "print\"Magnetic field at a point midway between two magnets is\",round(B*10**2,3),\"10**-2 T\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Magnetic field at a point midway between two magnets is 2.236 10**-2 T\n"
- ]
- }
- ],
- "prompt_number": 34
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 13.6 Page no 416"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "B=0.3 #T\n",
- "A=30 #Degree\n",
- "t=0.06 #Nn\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "M=t/(B*math.sin(A*3.14/180.0))\n",
- "\n",
- "#Result\n",
- "print\"Magnetic moment is\",round(M,1),\"Am**2\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Magnetic moment is 0.4 Am**2\n"
- ]
- }
- ],
- "prompt_number": 38
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 13.7 Page no 416"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "m=48 #Am\n",
- "l=0.25 #m\n",
- "a=30 #degree\n",
- "B=0.15\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "M=m*l\n",
- "T=M*B*math.sin(a*3.14/180.0)\n",
- "\n",
- "#Result \n",
- "print\"(i) Magnetic moment is\",M,\"Am**2\"\n",
- "print\"(ii) Torque is\",round(T,1),\"Mm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(i) Magnetic moment is 12.0 Am**2\n",
- "(ii) Torque is 0.9 Mm\n"
- ]
- }
- ],
- "prompt_number": 45
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 13.8 Page no 416"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "B=0.2 #t\n",
- "a=30\n",
- "t=0.06\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "M=t/(B*math.sin(a*3.14/180.0))\n",
- "\n",
- "#Result\n",
- "print\"(a) Magnetic moment is\",round(M,1),\"An**2\"\n",
- "print\"(b) When the magnet aligns itself parallel to the magnetic field\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a) Magnetic moment is 0.6 An**2\n",
- "(b) When the magnet aligns itself parallel to the magnetic field\n"
- ]
- }
- ],
- "prompt_number": 50
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 13.9 Page no 416"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "M=1.5 #J/T\n",
- "B=0.22 #T\n",
- "a=0\n",
- "a1=90\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "W=M*B*(math.cos(a*3.14/180.0)-math.cos(a1*3.14/180.0))\n",
- "t=M*B*math.sin(a1*3.14/180.0)\n",
- "\n",
- "#Result\n",
- "print\"Amount of work required is\", round(W,2),\"J\"\n",
- "print\"Torque is\",round(t,2),\"Nm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Amount of work required is 0.33 J\n",
- "Torque is 0.33 Nm\n"
- ]
- }
- ],
- "prompt_number": 57
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 13.10 Page no 417"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "n=50\n",
- "a=0.2 \n",
- "I=12 #A\n",
- "u=10**-7\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "B=u*2*math.pi*n*I/a\n",
- "M=n*I*math.pi*a**2\n",
- "\n",
- "#Result\n",
- "print\"(a) Magnetic moment at the centre is\",round(B*10**3,3),\"*10**-3 T\"\n",
- "print\"(b) Magnetic moment associated is\",round(M,1),\"Am**2\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a) Magnetic moment at the centre is 1.885 *10**-3 T\n",
- "(b) Magnetic moment associated is 75.4 Am**2\n"
- ]
- }
- ],
- "prompt_number": 65
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 13.11 Page no 417"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "r=0.53*10**-10 #m\n",
- "v=6.8*10**15 #Kz\n",
- "e=1.6*10**-19 #C\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "M=math.pi*e*v*r**2\n",
- "\n",
- "#Result\n",
- "print\"Magnetic moment is\",round(M*10**24,1)*10**-24,\"A m**2\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Magnetic moment is 9.6e-24 A m**2\n"
- ]
- }
- ],
- "prompt_number": 69
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 13.12 Page no 417"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "M=1.8*10**-23 #Am**2\n",
- "v=5*10**-6 #m**3\n",
- "a=7.8*10**3\n",
- "N=6.02*10**26\n",
- "A=56.0\n",
- "B=1.5 #T\n",
- "a1=90 #degree\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "m=v*a\n",
- "n=N*m/A\n",
- "M1=M*n\n",
- "t=M1*B*math.sin(a1*3.14/180.0)\n",
- "\n",
- "#Result\n",
- "print\"(a) Magnetic moment is\", round(M1,2),\"A m**2\"\n",
- "print\"(b) Torque required is\",round(t,2),\"N m\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a) Magnetic moment is 7.55 A m**2\n",
- "(b) Torque required is 11.32 N m\n"
- ]
- }
- ],
- "prompt_number": 79
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 13.13 Page no 417"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "B1=1.2*10**-2 #T\n",
- "a=15 #degree\n",
- "a2=60\n",
- "\n",
- "#Calculation\n",
- "a3=a2-a\n",
- "B2=B1*math.sin(a*3.14/180.0)/(math.sin(a3*3.14/180.0))\n",
- "\n",
- "#Result\n",
- "print round(B2*10**3,2),\"*10**-3 T\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "4.39 *10**-3 T\n"
- ]
- }
- ],
- "prompt_number": 83
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 13.14 Page no 417"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "n=1000\n",
- "I=2 #A\n",
- "A=2*10**-4 #m**@\n",
- "a=30\n",
- "B=0.16\n",
- "a1=0 #degree\n",
- "a2=180 \n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "M=I*A*n\n",
- "t=M*B*math.sin(a*3.14/180.0)\n",
- "W=M*B*(math.cos(a1*3.14/180.0)-math.cos(a2*3.14/180.0))\n",
- "\n",
- "#Result\n",
- "print\"(a) Torque required is\", round(t,3),\"Nm\"\n",
- "print\"(b) Work needed is\",round(W,3),\"J\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a) Torque required is 0.032 Nm\n",
- "(b) Work needed is 0.128 J\n"
- ]
- }
- ],
- "prompt_number": 92
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/modern_physics_by_Satish_K._Gupta/chap14.ipynb b/modern_physics_by_Satish_K._Gupta/chap14.ipynb deleted file mode 100755 index 9106ddd9..00000000 --- a/modern_physics_by_Satish_K._Gupta/chap14.ipynb +++ /dev/null @@ -1,635 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:04d8806c560b3bd07d87813351e39839b8f2c2b391de8dcf1afa3dc83f2be11d"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 14 Earths magnetism"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 14.3 Page no 433"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "a=60 #Degree\n",
- "Bh=0.16 #G\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "B=Bh/cos(a*3.14/180.0)\n",
- "\n",
- "#Result\n",
- "print\"Magnitude of earth's field is\", round(B,2),\"G\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Magnitude of earth's field is 0.32 G\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 14.4 Page no 434"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "a=60\n",
- "a2=45\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "a1=math.tan(a2*3.14/180.0)/math.cos(a*3.14/180.0)\n",
- "a3=math.atan(a1)*180/3.14\n",
- "\n",
- "#Result\n",
- "print\"Apparent value of the dip is\", round(a3,1),\"degree\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Apparent value of the dip is 63.4 degree\n"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 14.6 Page no 434"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "l=30 #cm\n",
- "l=0.15 #m\n",
- "r=0.30 #m\n",
- "Bh=0.34*10**-4 #T\n",
- "u=10**-7\n",
- "\n",
- "#Calculation\n",
- "M=Bh*(r**2-l**2)**2/(2*u*r)\n",
- "m=M/(2*l)\n",
- "\n",
- "#Result\n",
- "print\"Pole strength of the magnet is\",m,\"Am\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Pole strength of the magnet is 8.60625 Am\n"
- ]
- }
- ],
- "prompt_number": 16
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 14.7 Page no 434"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "M=0.4 #Am**2\n",
- "r=0.1 #m\n",
- "l=0.05 #m\n",
- "u=10**-7\n",
- "\n",
- "#Calculation\n",
- "Bh=u*M/((r**2+l**2)**1.5)\n",
- "\n",
- "#Result\n",
- "print\"Horizontal component of earth's magnetic field is\", round(Bh*10**4,3)*10**-4,\"T\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Horizontal component of earth's magnetic field is 2.86e-05 T\n"
- ]
- }
- ],
- "prompt_number": 22
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 14.8 Page no 435"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "B=0.33\n",
- "a=0\n",
- "u=10**-7\n",
- "I=2.5 #A\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "Bh=B/math.cos(a*3.14/180.0)\n",
- "a=u*2*I/(Bh*10**-4)\n",
- "\n",
- "#Result\n",
- "print\"Neutral point is\", round(a*10**2,1),\"cm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Neutral point is 1.5 cm\n"
- ]
- }
- ],
- "prompt_number": 28
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 14.9 Page no 435"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "Bh=0.32 #G\n",
- "B=0.48 \n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "a=B/Bh\n",
- "a1=math.atan(a)*180/3.14\n",
- "\n",
- "#Result\n",
- "print\"New stable equilibrium is\", round(a1,1),\"degree\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "New stable equilibrium is 56.3 degree\n"
- ]
- }
- ],
- "prompt_number": 32
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 4.10 Page no 435"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "n=22\n",
- "a=0.1 #m\n",
- "Bh=0.3*10**-4 #T\n",
- "u=10**-7\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "K=2*a*Bh/(n*4*math.pi*u)\n",
- "\n",
- "#Result\n",
- "print\"Reduction factor is\", round(K,3),\"A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Reduction factor is 0.217 A\n"
- ]
- }
- ],
- "prompt_number": 36
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 14.11 Page no 435"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "n=40\n",
- "a=0.12\n",
- "I=0.15\n",
- "a1=45 #degree\n",
- "u=10**-7\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "Bh=(n*u*4*math.pi*I)/(2.0*a*math.tan(a1*3.14/180.0))\n",
- "\n",
- "#Result\n",
- "print\"Strength of horizontal component is\", round(Bh*10**4,3),\"*10**-4 T\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Strength of horizontal component is 0.314 *10**-4 T\n"
- ]
- }
- ],
- "prompt_number": 51
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 14.12 Page no 435"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "a1=30\n",
- "a2=45 #degree\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "n=2*math.tan(a1*3.14/180.0)/(math.tan(a2*3.14/180.0))\n",
- "\n",
- "#Result\n",
- "print\"Ratio of number of turns is\", round(n,3)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Ratio of number of turns is 1.155\n"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 14.13 Page no 436"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "n=16\n",
- "a=0.1 #m\n",
- "i=0.75 #A\n",
- "Bh=5*10**-2 #T\n",
- "v=2 #/s\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "M=n*i*math.pi*a**2\n",
- "I=M*Bh/(4*math.pi**2*v**2)\n",
- "\n",
- "#Result\n",
- "print\"Moment of inertia is\",round(I*10**4,3),\"*10**-4 Kg m**2\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Moment of inertia is 1.194 *10**-4 Kg m**2\n"
- ]
- }
- ],
- "prompt_number": 10
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 14.14 Page no 436"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "T1=2.5\n",
- "T2=4.5\n",
- "M2=1.5\n",
- "\n",
- "#Calculation\n",
- "M=T2**2/(M2*T1**2)\n",
- "\n",
- "#Result\n",
- "print\"Ratio of magnetic moment is\",M"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Ratio of magnetic moment is 2.16\n"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 14.15 Page no 436"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "T1=3.0\n",
- "T2=4.0\n",
- "\n",
- "#Calculation\n",
- "M=(T2**2+T1**2)/(T2**2-T1**2)\n",
- "\n",
- "#Result\n",
- "print\"Ratio of magnetic moments is\",round(M,2)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Ratio of magnetic moments is 3.57\n"
- ]
- }
- ],
- "prompt_number": 15
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 14.16 Page no 436"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "a1=35 #Degree\n",
- "B=0.39\n",
- "I=1 #A\n",
- "a=4.0*10**-2\n",
- "u=10**-7\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "Bh=B*math.cos(a1*3.14/180.0)\n",
- "Bv=B*math.sin(a1*3.14/180.0)\n",
- "B1=(u*2*I*4/a)*10**4\n",
- "Rh=Bh-B1\n",
- "R=math.sqrt(Rh**2+Bv**2)\n",
- "Rh1=Bh+B1\n",
- "R3=math.sqrt(Rh1**2+Bv**2)\n",
- "\n",
- "#Result\n",
- "print\"Resultant magnetic field is\", round(R3,3),\"G\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Resultant magnetic field is 0.566 G\n"
- ]
- }
- ],
- "prompt_number": 39
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 14.17 Page no 436"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "M=5.25*10**-2 #J/T\n",
- "Bh=0.42*10**-4 #T\n",
- "u=10**-7\n",
- "\n",
- "#Calculation\n",
- "r=(u*M/Bh)**0.333\n",
- "r1=(u*2*M/Bh)**0.333\n",
- "\n",
- "#Result\n",
- "print\"(a) Distance from the centre of the magnet on its normal bisector is\", round(r*10**2,1),\"cm\"\n",
- "print\"(b) Distance from the centre of the magnet on its axis is\",round(r1*10**2,1),\"cm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a) Distance from the centre of the magnet on its normal bisector is 5.0 cm\n",
- "(b) Distance from the centre of the magnet on its axis is 6.3 cm\n"
- ]
- }
- ],
- "prompt_number": 51
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 14.18 Page no 437"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "I=0.35 #A\n",
- "n=30\n",
- "a=12.0*10**-2\n",
- "u=10**-7\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "Bh=u*2*math.pi*n*I*0.707/a\n",
- "\n",
- "#Result\n",
- "print\"(a) Horizontal component of the earth's magnetic field is\", round(Bh*10**4,2),\"G\"\n",
- "print\"(b) The needle will reverse its original direction i.e. it will point east to west.\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a) Horizontal component of the earth's magnetic field is 0.39 G\n",
- "(b) The needle will reverse its original direction i.e. it will point east to west.\n"
- ]
- }
- ],
- "prompt_number": 60
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 14.19 Page no 437"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#given\n",
- "t1=9 #S\n",
- "t2=4.5\n",
- "Bh=0.34*10**-4 #T\n",
- "u=10**-7\n",
- "r=0.1 #m\n",
- "\n",
- "#Calculation\n",
- "Baxial=Bh*((t1**2/t2**2)-1)\n",
- "M2=Baxial*r**2/(2*u)\n",
- "\n",
- "#Result\n",
- "print\"Magnetic moment is\", M2*10**-1,\"A m**2\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Magnetic moment is 0.51 A m**2\n"
- ]
- }
- ],
- "prompt_number": 66
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/modern_physics_by_Satish_K._Gupta/chap15.ipynb b/modern_physics_by_Satish_K._Gupta/chap15.ipynb deleted file mode 100755 index f90ea520..00000000 --- a/modern_physics_by_Satish_K._Gupta/chap15.ipynb +++ /dev/null @@ -1,317 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:859464a7b9c1dff7d24f65e6024e51cc8bc55fc7979a3e8c04919720ca49d3b9"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 15 Classification of Magnetic Materials"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 15.1 Page no 457"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "q=0.12 #TA**-1m\n",
- "q1=4*math.pi*10**-7 #TA**-1m\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "q3=q/q1\n",
- "e=q3-1\n",
- "\n",
- "#Result\n",
- "print\"Relative permeability is\",round(q3*10**-4,2),\"10**4\"\n",
- "print\"susceptibility is\",round(e*10**-4,2),\"10**4\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Relative permeability is 9.55 10**4\n",
- "susceptibility is 9.55 10**4\n"
- ]
- }
- ],
- "prompt_number": 17
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 15.2 Page no 457"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "A=0.5*10**-4 #m**2\n",
- "H=1200 #Am**-1\n",
- "W=599\n",
- "Q=4*math.pi*10**-7 #TA\"**-1m\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "Z=Q*(1+W)\n",
- "Z1=Z*H\n",
- "Z2=Z1*A\n",
- "\n",
- "#Result\n",
- "print\"(i) Z produced is\",round(Z*10**4,2),\"10**-4\",\"TA**-1m\"\n",
- "print\"(ii) Z1 produced is\",round(Z1,3),\"T\"\n",
- "print\"(iii) Z2 produced is\",round(Z2*10**5,3),\"10**-5\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(i) Z produced is 7.54 10**-4 TA**-1m\n",
- "(ii) Z1 produced is 0.905 T\n",
- "(iii) Z2 produced is 4.524 10**-5\n"
- ]
- }
- ],
- "prompt_number": 66
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 15.3 Page no 457"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "I=1.2 #A\n",
- "Q=800\n",
- "A=0.15 #m\n",
- "S=3500\n",
- "K=4*math.pi*10**-7\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "L=2*math.pi*A\n",
- "n=S/L\n",
- "B=(K*Q*S*I)/L\n",
- "\n",
- "#Result\n",
- "print\"The magnetic field in the core for a magnetising current of 1.2 A is\",B,\"T\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The magnetic field in the core for a magnetising current of 1.2 A is 4.48 T\n"
- ]
- }
- ],
- "prompt_number": 73
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 15.4 Page no 457"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "H=2*10**3 #Am**-1\n",
- "I=4.8*10**-2 #Am**-1\n",
- "T=280 #K\n",
- "T1=320.0 #K\n",
- "\n",
- "#Calculation\n",
- "S=I/H\n",
- "S1=S*(T/T1)\n",
- "S2=S1*H\n",
- "\n",
- "#Result\n",
- "print\"The susceptibility of aluminium is raised to 280 k is\",S*10**5,\"10**3\"\n",
- "print\"The temperature of the aluminium is raised to 320 k is\",S1\n",
- "print\"The suscepitibility and intensity of magnetistionis\",S2*10**2,\"10**-2\",\"Am**-1\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The susceptibility of aluminium is raised to 280 k is 2.4 10**3\n",
- "The temperature of the aluminium is raised to 320 k is 2.1e-05\n",
- "The suscepitibility and intensity of magnetistionis 4.2 10**-2 Am**-1\n"
- ]
- }
- ],
- "prompt_number": 90
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 15.5 Page no 458"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "W=3.2*10**4 #J\n",
- "M=8.4\n",
- "D=7200\n",
- "v=50 #cycle s**-1\n",
- "T=30*60 #s\n",
- "\n",
- "#Calculation\n",
- "V=M/D\n",
- "Q=W/(V*v*T)\n",
- "\n",
- "#Result\n",
- "print\"The value of energy is\",round(Q,1),\"J m**-3 cycle**-1\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of energy is 304.8 J m**-3 cycle**-1\n"
- ]
- }
- ],
- "prompt_number": 95
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 15.6 Page no 458"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "m=4.5 #Am\n",
- "l=0.06 #m\n",
- "A=0.9*10**-4 #m**2\n",
- "r=0.06 #m\n",
- "J=4*math.pi*10**-7\n",
- "\n",
- "#Calculation\n",
- "I=m/A\n",
- "H1=m/(4*math.pi*r**2)\n",
- "H2=m/(4*math.pi*r**2)\n",
- "H=H1+H2\n",
- "B=J*(-H+I)\n",
- "\n",
- "#Result\n",
- "print\"(a) Intensity of the magnetisation is\", I*10**-4,\"10**4 A/m\"\n",
- "print\"(b) Magnetic intensity is\",round(H,0),\"A/m\"\n",
- "print\"(c) Magnetic induction is\",round(B*10**2,2),\"*10**-2 T\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a) Intensity of the magnetisation is 5.0 10**4 A/m\n",
- "(b) Magnetic intensity is 199.0 A/m\n",
- "(c) Magnetic induction is 6.26 *10**-2 T\n"
- ]
- }
- ],
- "prompt_number": 16
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 15.7 Page no 458"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "m=1.5*10**-23 #J/T\n",
- "n=2.0*10**24\n",
- "a=15 #%\n",
- "T1=4.2 #K\n",
- "T2=2.8\n",
- "B1=0.84 #T\n",
- "B2=0.98\n",
- "\n",
- "#Calculation\n",
- "M=m*n\n",
- "M1=M*(a/100.0)\n",
- "M2=M1*T1*B2/(T2*B1)\n",
- "\n",
- "#Result\n",
- "print\"Total dipole moment is\",M2,\"J/T\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Total dipole moment is 7.875 J/T\n"
- ]
- }
- ],
- "prompt_number": 18
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/modern_physics_by_Satish_K._Gupta/chap16.ipynb b/modern_physics_by_Satish_K._Gupta/chap16.ipynb deleted file mode 100755 index a6266584..00000000 --- a/modern_physics_by_Satish_K._Gupta/chap16.ipynb +++ /dev/null @@ -1,964 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:03e25d70b6f5039934ebff67f76c18265eda838017719c81d51174de65a0d0cb"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 16 Electromagnetic induction"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.1 Page no 508"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "R=5.0 #ohm\n",
- "t=2\n",
- "a=15\n",
- "b=8\n",
- "\n",
- "#Calculation\n",
- "e=-a*t**2-(b*t)-t\n",
- "I=-e/R\n",
- "\n",
- "#Result\n",
- "print\"Induced current is\", I,\"A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Induced current is 15.6 A\n"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.2 Page no 508"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "S1=75*10**-4 #m**2\n",
- "B1=0.8 #wb/m**2\n",
- "S2=100*10**-4\n",
- "B2=1.4\n",
- "t=0.05\n",
- "\n",
- "#Calculation\n",
- "a1=B1*S1\n",
- "a2=B2*S2\n",
- "a=a2-a1\n",
- "e=-a/t\n",
- "\n",
- "#Result\n",
- "print\"Induced e.m.f is\", e,\"Volt\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Induced e.m.f is -0.16 Volt\n"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.3 Page no 508"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "a1=5.5*10**-4 #Wb\n",
- "a2=0.5*10**-4\n",
- "N=1000\n",
- "t=0.1\n",
- "R=10 #ohm\n",
- "\n",
- "#Calculation\n",
- "a=a2-a1\n",
- "a11=N*a\n",
- "e=-(a11/t)\n",
- "I1=e/R\n",
- "I2=I1*t\n",
- "\n",
- "#Result\n",
- "print\" Induced e.m.f produced is\",e,\"V\"\n",
- "print\" Charge flowing through the coil in 0.1 is\",I2,\"C\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- " Induced e.m.f produced is 5.0 V\n",
- " Charge flowing through the coil in 0.1 is 0.05 C\n"
- ]
- }
- ],
- "prompt_number": 27
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.5 Page no 509"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "B=2.5*10**-3 #Wb**-2\n",
- "L=1 #m\n",
- "v=30 #r.p.s\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "e=-B*math.pi*L**2*v\n",
- "\n",
- "#Result\n",
- "print\"The produced e.m.f. between its ends is\",round(e,3),\"V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The produced e.m.f. between its ends is -0.236 V\n"
- ]
- }
- ],
- "prompt_number": 31
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.6 Page no 509"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "L=1.2 #m\n",
- "e=10**-2 #v\n",
- "B=5*10**-5 #tesla\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "V=e/(B*math.pi*L**2)\n",
- "\n",
- "#Result\n",
- "print\"The rate of rotation of the wheel is\",round(V,1),\"Rotation a**-1\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The rate of rotation of the wheel is 44.2 Rotation a**-1\n"
- ]
- }
- ],
- "prompt_number": 37
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.7 Page no 509"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "D=10\n",
- "L=0.50 #m\n",
- "B=0.40*10**-4 #T\n",
- "V=2 #r.p.s.\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "E=-B*math.pi*L**2*V\n",
- "\n",
- "#Result\n",
- "print\"The induced e.m.f. between the axle and the rim of the wheel is\",round(E*10**5,3),\"10**-5\",\"V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The induced e.m.f. between the axle and the rim of the wheel is -6.283 10**-5 V\n"
- ]
- }
- ],
- "prompt_number": 45
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.8 Page no 509"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "B=0.2 #T\n",
- "r=0.1 #m\n",
- "R=2 #ohm\n",
- "D=20*math.pi #rad s**-1\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "V=D/(2*math.pi)\n",
- "E=-B*math.pi*r**2*V\n",
- "I=E/R\n",
- "\n",
- "#Result\n",
- "print\"(i) The potential difference is\",round(E,4),\"V\"\n",
- "print\"(ii) The induced current is\",round(I,4),\"A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(i) The potential difference is -0.0628 V\n",
- "(ii) The induced current is -0.0314 A\n"
- ]
- }
- ],
- "prompt_number": 56
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.9 Page no 509"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "B=8.0*10**-5 #Wb m**-2\n",
- "L=2 #m\n",
- "v=30 #m s**-1\n",
- "\n",
- "#Calculation\n",
- "e=B*L*v\n",
- "\n",
- "#Result\n",
- "print\"The vertical component of earth's field is\",e*10**3,\"10**-3\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The vertical component of earth's field is 4.8 10**-3\n"
- ]
- }
- ],
- "prompt_number": 61
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.10 Page no 509"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Givem\n",
- "l=10 #m\n",
- "v=5 #m/s\n",
- "Bh=0.30*10**-4 #Wb/m**2\n",
- "\n",
- "#Calculation\n",
- "e=Bh*l*v\n",
- "\n",
- "#Result\n",
- "print \"Instantaneous value of e.m.f. induced is\",e*10**3,\"*10**-3 V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Instantaneous value of e.m.f. induced is 1.5 *10**-3 V\n"
- ]
- }
- ],
- "prompt_number": 64
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.11 Page no 509"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "B=0.3 #T\n",
- "v=10**-2 #m/s\n",
- "l=8*10**-2\n",
- "L=1\n",
- "v1=1.0\n",
- "l2=2*10**-2\n",
- "L1=8\n",
- "\n",
- "#Calculation\n",
- "e=B*l*v\n",
- "l=L/v1\n",
- "e1=B*l2*v\n",
- "t1=L1/v1\n",
- "\n",
- "#Result\n",
- "print\"(i) Voltage developed in the direction of motion normal to the longer side is\", e*10**3,\"mv\"\n",
- "print\"(ii) Voltage developed in the direction of motion normal to the shorter side is\",e1*10**3,\"mV\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(i) Voltage developed in the direction of motion normal to the longer side is 0.24 mv\n",
- "(ii) Voltage developed in the direction of motion normal to the shorter side is 0.06 mV\n"
- ]
- }
- ],
- "prompt_number": 72
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.12 Page no 510"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "B=0.4 #T\n",
- "v=5 #m/s\n",
- "l=0.25 #m\n",
- "e=0.5\n",
- "R=5.0\n",
- "\n",
- "#Calculation\n",
- "e=B*l*v\n",
- "I=e/R\n",
- "\n",
- "#Result\n",
- "print\"New current is\",I,\"A\"\n",
- "print\"Direction is from the end S to R\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "New current is 0.1 A\n",
- "Direction is from the end S to R\n"
- ]
- }
- ],
- "prompt_number": 75
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.13 Page no 510"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "v=500 #m/s\n",
- "B=5*10**-4 #t\n",
- "a=30 #degree\n",
- "l=25 #m\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "Bv=B*math.sin(a*3.14/180.0)\n",
- "e=Bv*l*v\n",
- "\n",
- "#Result\n",
- "print\"Voltage difference is\", round(e,3),\"V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Voltage difference is 3.124 V\n"
- ]
- }
- ],
- "prompt_number": 80
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.14 Page no 510"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "L=10*10**-3 #H\n",
- "I=4*10**-3 #A\n",
- "N=200.0\n",
- "\n",
- "#Calculation\n",
- "a=L*I\n",
- "A=a/N\n",
- "\n",
- "#Result\n",
- "print\"Total magnetic flux is\",a,\"Weber\"\n",
- "print\"Magnetic flux through the cross section is\",A,\"Weber\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Total magnetic flux is 4e-05 Weber\n",
- "Magnetic flux through the cross section is 2e-07 Weber\n"
- ]
- }
- ],
- "prompt_number": 85
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.15 Page no 510"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "L=10**-2 #H\n",
- "I1=0\n",
- "I2=1 #A\n",
- "l=1 \n",
- "t=0.01\n",
- "\n",
- "#Calculation\n",
- "e=-L*(l/t)\n",
- "\n",
- "#Result\n",
- "print\"Self induced e.m.f is\", e,\"V\"\n",
- "print\"The self-induced e.m.f. will act so as to oppose the growth of current.\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Self induced e.m.f is -1.0 V\n",
- "The self-induced e.m.f. will act so as to oppose the growth of current.\n"
- ]
- }
- ],
- "prompt_number": 89
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.16 Page no 510"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "n=1500 #turns/m\n",
- "A=2*10**-4 #m**2\n",
- "l=20 #A/s\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "e=-4*math.pi*n*A*l*10**-7\n",
- "\n",
- "#Result\n",
- "print\"Induced e.m.f. is\",round(e*10**6,2)*10**-6,\"V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Induced e.m.f. is -7.54e-06 V\n"
- ]
- }
- ],
- "prompt_number": 97
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.17 Page no 511"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "e=50*10**-3 #V\n",
- "a=8\n",
- "b=4\n",
- "t=0.5\n",
- "\n",
- "#Calculation\n",
- "l=a-b\n",
- "M=e*t/l\n",
- "\n",
- "#Result\n",
- "print\"Mutual inductance is\",M*10**3,\"*10**-3 H\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Mutual inductance is 6.25 *10**-3 H\n"
- ]
- }
- ],
- "prompt_number": 101
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.18 Page no 511"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "n1=5000 #turns/m\n",
- "A=4*10**-4 #m**2\n",
- "n2l=200\n",
- "u=10**-7\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "M=4*math.pi*u*n1*n2l*A\n",
- "\n",
- "#Result\n",
- "print\"Mutual inductance is\", round(M*10**4,3),\"*10**-4 H\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Mutual inductance is 5.027 *10**-4 H\n"
- ]
- }
- ],
- "prompt_number": 109
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.19 Page no 511"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "N=1200\n",
- "A=12*10**-4\n",
- "r=15*10**-2 #m\n",
- "u=10**-7\n",
- "I=1.0\n",
- "N2=300\n",
- "I1=0\n",
- "I2=2\n",
- "t=0.05\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "n=N/(2*math.pi*r)\n",
- "B=4*math.pi*u*n\n",
- "a=B*A*N\n",
- "L=a/I\n",
- "a1=B*A*N2\n",
- "a11=a1*I1\n",
- "a12=a1*I2\n",
- "a13=a12-a11\n",
- "e=-a13/t\n",
- "\n",
- "#Result\n",
- "print\"(a) Self inductance is\", L*10**3,\"*10**-3 H\"\n",
- "print\"(b) Induced e.m.f is\", round(e,3),\"V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a) Self inductance is 2.304 *10**-3 H\n",
- "(b) Induced e.m.f is -0.023 V\n"
- ]
- }
- ],
- "prompt_number": 124
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.20 Page no 511"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "A=25*10**-4\n",
- "N=500\n",
- "l=30.0*10**-2 #m\n",
- "I=2.5\n",
- "u=10**-7\n",
- "t=10.0**-3\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "n=N/l\n",
- "B=4*math.pi*u*n*I\n",
- "a=B*A*N\n",
- "a1=0-a\n",
- "e=-a1/t\n",
- "\n",
- "#Result\n",
- "print\"Average induced e.m.f. produced is\", round(e,3),\"V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Average induced e.m.f. produced is 6.545 V\n"
- ]
- }
- ],
- "prompt_number": 130
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.21 Page no 511"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "N=25.0\n",
- "A=2*10**-4 #m**2\n",
- "q=7.5*10**-3\n",
- "R=0.50\n",
- "\n",
- "#Calculation\n",
- "B=R*q/(N*A)\n",
- "\n",
- "#Result\n",
- "print\"Field strength of the magnet is\" ,B,\"T\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Field strength of the magnet is 0.75 T\n"
- ]
- }
- ],
- "prompt_number": 136
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.22 Page no 512"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "B=0.50 #T\n",
- "l=15*10**-2 #m\n",
- "R=9.0*10**-3 #ohm\n",
- "v=12*10**-2 #m/s\n",
- "\n",
- "#Calculation\n",
- "e=B*v*l\n",
- "F=B*l*(e/R)\n",
- "P=F*v\n",
- "P1=e**2/R\n",
- "\n",
- "#Result\n",
- "print\"(a) Induced e.m.f is\", e*10**3,\"*10**3 V\"\n",
- "print\"The end P of the rod will become positive and the end Q will become negative\"\n",
- "print\"(b) on closing the switch, electrons collects at the end Q. Therefore , excess charge is built up i.e it doesn't open when switch is open\" \n",
- "print\"(c) The magnetic lorentz force on electron is cancelled due to the electric field set up across the two end\"\n",
- "print\"(d) Retarding force is\",F*10**2,\"*10**-2 N\"\n",
- "print\"(e) Power is\",P*10**3,\"*10*-3 W\"\n",
- "print\"(f) Dissipated power is\",P*10**3,\"*10**-3 W\"\n",
- "print\"(g) The motion of the rod does not cut field lines,hence no induced e.m.f. is produced\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a) Induced e.m.f is 9.0 *10**3 V\n",
- "The end P of the rod will become positive and the end Q will become negative\n",
- "(b) on closing the switch, electrons collects at the end Q. Therefore , excess charge is built up i.e it doesn't open when switch is open\n",
- "(c) The magnetic lorentz force on electron is cancelled due to the electric field set up across the two end\n",
- "(d) Retarding force is 7.5 *10**-2 N\n",
- "(e) Power is 9.0 *10*-3 W\n",
- "(f) Dissipated power is 9.0 *10**-3 W\n",
- "(g) The motion of the rod does not cut field lines,hence no induced e.m.f. is produced\n"
- ]
- }
- ],
- "prompt_number": 156
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.23 Page no 512"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "a1=20*10**-2\n",
- "a2=0.3*10**-2 #m\n",
- "x=15*10**-2\n",
- "I=2.0 #A\n",
- "u=10**-7\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "B1=u*2*math.pi*I*a1**2/((a1**2+x**2)**1.5)\n",
- "a=B1*math.pi*a2**2\n",
- "M=a/I\n",
- "\n",
- "#Result\n",
- "print\"(a) Flux is\", round(B1*10**6,3)*10**-6 ,\"T\"\n",
- "print\"(b) Mutual inductance is\",round(M*10**11,3)*10**-11,\"H\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a) Flux is 3.217e-06 T\n",
- "(b) Mutual inductance is 4.548e-11 H\n"
- ]
- }
- ],
- "prompt_number": 168
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.25 Page no 513"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "N1=1500\n",
- "l1=80.0*10**-2\n",
- "l2=4*10**-2\n",
- "r2=2*10**-2\n",
- "I=3.0 #A\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "A2=math.pi*r**2\n",
- "n1=N1/l1\n",
- "a=4*math.pi*10**-7*n1*I*4*math.pi*10**-4*100\n",
- "M=a/I\n",
- "\n",
- "#Result\n",
- "print\"Flux is\",round( a*10**4,3),\"*10**-4 Wb\"\n",
- "print\"Mutual inductance is\",round(M*10**4,2),\"*10**-4 H\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Flux is 8.883 *10**-4 Wb\n",
- "Mutual inductance is 2.96 *10**-4 H\n"
- ]
- }
- ],
- "prompt_number": 181
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/modern_physics_by_Satish_K._Gupta/chap17.ipynb b/modern_physics_by_Satish_K._Gupta/chap17.ipynb deleted file mode 100755 index cb448cf8..00000000 --- a/modern_physics_by_Satish_K._Gupta/chap17.ipynb +++ /dev/null @@ -1,263 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:7c5cc6b58c18456d76e825a85df242d83f70c07f62b0e167355c9552ee60085c"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 17 Transient Current"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.1 Page no 534"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "E=3 #V\n",
- "L=2.5 #H\n",
- "R=50.0 #ohm\n",
- "\n",
- "#Calculation\n",
- "T=L/R\n",
- "L1=E/L\n",
- "I0=E/R\n",
- "\n",
- "#Result\n",
- "print\"(i) Time constant is\", T,\"S\"\n",
- "print\"(ii) Rate of change of current is\",L1,\"A/S\"\n",
- "print\"(iii) Maximum current is\",I0,\"A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(i) Time constant is 0.05 S\n",
- "(ii) Rate of change of current is 1.2 A/S\n",
- "(iii) Maximum current is 0.06 A\n"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.2 Page no 534"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "E=1 #V\n",
- "R=1 #ohm\n",
- "L=1 #H\n",
- "t=1 #S\n",
- "a=2.3036\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "I=1-math.exp(-E/L)\n",
- "t=a*math.log10(2)\n",
- "\n",
- "#Result\n",
- "print\"(i) Current after 1 S is\", round(I,3),\"A\"\n",
- "print\"(ii) Time to take the current to reach half its final value is\",round(t,3),\"S\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(i) Current after 1 S is 0.632 A\n",
- "(ii) Time to take the current to reach half its final value is 0.693 S\n"
- ]
- }
- ],
- "prompt_number": 15
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.3 Page no 534"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "I0=2 #A\n",
- "L=0.1 #H\n",
- "R=20 #ohm\n",
- "I=0.3 \n",
- "a=3.333\n",
- "b=200.0\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "t=2.3026*math.log10(a)/b\n",
- "l=R*I*I0/L\n",
- "\n",
- "#Result\n",
- "print\"Rate of change of current is\", l,\"A/s\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Rate of change of current is 120.0 A/s\n"
- ]
- }
- ],
- "prompt_number": 19
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.4 Page no 535"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "R=2*10**6 #ohm\n",
- "C=10**-6 #farad\n",
- "q=0.8647\n",
- "a=0.1353\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "t=-2*2.3026*math.log10(a)\n",
- "\n",
- "#Result\n",
- "print\"Time is\", round(t,0),\"S\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Time is 4.0 S\n"
- ]
- }
- ],
- "prompt_number": 29
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.5 Page no 535"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "C=1.443*10**-6 #F\n",
- "t=60 #S\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "a=2.303*math.log10(2)\n",
- "R=t/(C*a)\n",
- "\n",
- "#Result\n",
- "print\"Value of resistance is\", round(R*10**-6,0),\"*10**6 ohm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Value of resistance is 60.0 *10**6 ohm\n"
- ]
- }
- ],
- "prompt_number": 36
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 17.6 Page no 535"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "L=200*10**-6 #H\n",
- "f1=8*10**5\n",
- "f2=12*10**-5 #Hz\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "C1=1/(4*math.pi**2*f1**2*L)\n",
- "C=1/(4*math.pi**2*f2**2*L)\n",
- "\n",
- "#Result\n",
- "print\"Range of capacitor C is\",round(C*10**-8,0),\"pF\"\n",
- "print\"Range of capacitor C1 is\",round(C1*10**12,0),\"pF\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Range of capacitor C is 88.0 pF\n",
- "Range of capacitor C1 is 198.0 pF\n"
- ]
- }
- ],
- "prompt_number": 52
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/modern_physics_by_Satish_K._Gupta/chap18.ipynb b/modern_physics_by_Satish_K._Gupta/chap18.ipynb deleted file mode 100755 index 94dff4f2..00000000 --- a/modern_physics_by_Satish_K._Gupta/chap18.ipynb +++ /dev/null @@ -1,1298 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:14dd6f2dca7ff6e4458a0a97e28043f8703a45199dee1a45b482cebdf3ee0a74"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 18 Alternating current"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.2 Page no 555"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "f1=100\n",
- "f2=2.0\n",
- "I0=50\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "f=f1/f2\n",
- "Im=0.636*I0\n",
- "Iv=0.707*I0\n",
- "I=I0*math.sin(60*3.14/180.0)\n",
- "\n",
- "#Result\n",
- "print\"(i) Frequency of A.C applied is\",f,\"c.p.s\"\n",
- "print\"(ii) Mean value of current is\",Im,\"A\"\n",
- "print\"(iii) Virtual value of current is\",Iv,\"A\"\n",
- "print\"(iv) Value of current is\",round(I,1),\"A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(i) Frequency of A.C applied is 50.0 c.p.s\n",
- "(ii) Mean value of current is 31.8 A\n",
- "(iii) Virtual value of current is 35.35 A\n",
- "(iv) Value of current is 43.3 A\n"
- ]
- }
- ],
- "prompt_number": 10
- },
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Example 18.3 Page no 555"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "R=10.0 #ohm\n",
- "E0=200\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "Ev=E0/math.sqrt(2)\n",
- "Iv=Ev/R\n",
- "Pav=Ev*Iv\n",
- "\n",
- "#Result\n",
- "print\"(i) r.m.s value of voltage is\", E0,\"V\"\n",
- "print\"(ii) r.m.s value of current is\",round(Iv,2),\"A\"\n",
- "print\"(iii)Power dissipated is\",Pav,\"W\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(i) r.m.s value of voltage is 200 V\n",
- "(ii) r.m.s value of current is 14.14 A\n",
- "(iii)Power dissipated is 2000.0 W\n"
- ]
- }
- ],
- "prompt_number": 18
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.4 Page no 555"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "L=1 #H\n",
- "Ev=110 #V\n",
- "f=70 #Hz\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "Xl=2*3.14*f*L\n",
- "I=Ev/Xl\n",
- "I0=math.sqrt(2)*I\n",
- "\n",
- "#Result\n",
- "print\"(a) Reactance is\", round(Xl,0)\n",
- "print\"(b) Current through inductance is\",round(I,2),\"A\"\n",
- "print\"(c) Peak value of current is\",round(I0,3),\"A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a) Reactance is 440.0\n",
- "(b) Current through inductance is 0.25 A\n",
- "(c) Peak value of current is 0.354 A\n"
- ]
- }
- ],
- "prompt_number": 28
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.5 Page no 555"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "L=2*10**-3\n",
- "w=200 #rad/s\n",
- "I0=0.2\n",
- "\n",
- "#Calculation\n",
- "Xl=w*L\n",
- "e=L*I0*w\n",
- "I0=e/Xl\n",
- "\n",
- "#Result\n",
- "print\"Maximum value of induced current is\", I0,\"A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Maximum value of induced current is 0.2 A\n"
- ]
- }
- ],
- "prompt_number": 35
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.6 Page no 555"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "C=10*10**-6 #F\n",
- "f=50 #cycles/s\n",
- "Ev=110\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "Xc=1/(math.pi*2*f*C)\n",
- "Iv=Ev/Xc\n",
- "\n",
- "#Result\n",
- "print\"Virtual value of current is\", round(Iv,3),\"A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Virtual value of current is 0.346 A\n"
- ]
- }
- ],
- "prompt_number": 40
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.7 Page no 555"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "L=4\n",
- "R=30 #ohm\n",
- "Ev=200\n",
- "f=50\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "Xl=2*math.pi*f*(L/math.pi)\n",
- "Z=math.sqrt(R**2+Xl**2)\n",
- "Iv=Ev/Z\n",
- "\n",
- "#Result\n",
- "print\"current flowing in the circuit is\", round(Iv,3),\"A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "current flowing in the circuit is 0.499 A\n"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.8 Page no 556"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "E=100\n",
- "I=1.0\n",
- "Iv=0.5\n",
- "f=50\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "R=E/I\n",
- "Z=E/Iv\n",
- "Xl=math.sqrt(Z**2-R**2)\n",
- "L=Xl/(2*math.pi*f)\n",
- "\n",
- "#Result\n",
- "print\"Inductance of the coil is\", round(L,2),\"H\"\n",
- "print\"Resistance is\",R,\"ohm\"\n",
- "print\"Impedence is\",Z,\"ohm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Inductance of the coil is 0.55 H\n",
- "Resistance is 100.0 ohm\n",
- "Impedence is 200.0 ohm\n"
- ]
- }
- ],
- "prompt_number": 18
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.9 Page no 556"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "L=0.50 #H\n",
- "R=100.0 #ohm\n",
- "f=50 #Hz\n",
- "Ev=240\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "Iv=Ev/(math.sqrt(R**2+(2*math.pi*f*L)**2))\n",
- "I0=math.sqrt(2)*Iv\n",
- "a=2*math.pi*f*L/R\n",
- "a1=math.atan(a)*180/3.14\n",
- "\n",
- "#Result\n",
- "print\"(a) Maximum current in the coil is\", round(I0,3),\"A\"\n",
- "print\"(b) Phase difference is\",round(a1,1),\"degree (e.m.f. leads current)\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a) Maximum current in the coil is 1.823 A\n",
- "(b) Phase difference is 57.5 degree (e.m.f. leads current)\n"
- ]
- }
- ],
- "prompt_number": 29
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.10 Page no 556"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "Ev=220\n",
- "Iv=0.5\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "R=Ev/Iv\n",
- "Xl=Ev/Iv\n",
- "Z=math.sqrt(R**2+Xl**2)\n",
- "Iv1=Ev/Z\n",
- "\n",
- "#Result\n",
- "print\"(a) X is a resistor of\",R,\"ohm\\n and Y is a inductor of\",Xl,\"ohm\"\n",
- "print\"(b) Current in the circuit is\",round(Iv1,3),\"A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a) X is a resistor of 440.0 ohm\n",
- " and Y is a inductor of 440.0 ohm\n",
- "(b) Current in the circuit is 0.354 A\n"
- ]
- }
- ],
- "prompt_number": 37
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.11 Page no 556"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "Z=100 #ohm\n",
- "a=45 #degree\n",
- "f=1000\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "Xl=Z/math.sqrt(2)\n",
- "L=Xl/(2*math.pi*f)\n",
- "\n",
- "#Result\n",
- "print\"Self inductance of the coil is\", round(L*10**2,4),\"*10**-2 H\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Self inductance of the coil is 1.1254 *10**-2 H\n"
- ]
- }
- ],
- "prompt_number": 43
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.12 Page no 556"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "R=10 #ohm\n",
- "Ev=220\n",
- "f=50 #Hz\n",
- "Iv=2.0 #A\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "Z=Ev/Iv\n",
- "Xc=math.sqrt(Z**2-R**2)\n",
- "\n",
- "#Result\n",
- "print\"Reactance of the capacitor is\",round(Xc,2),\"ohm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Reactance of the capacitor is 109.54 ohm\n"
- ]
- }
- ],
- "prompt_number": 46
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.13 Page no 556"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "R=10\n",
- "C=0.1*10**-6 #F\n",
- "Ev=100 #V\n",
- "f=50\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "Z=math.sqrt(R**2+(1/(2*math.pi*f*C))**2)\n",
- "Iv=Ev/Z\n",
- "\n",
- "#Result\n",
- "print\"Current in the circuit is\", round(Iv*10**3,3),\"*10**-3 A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Current in the circuit is 3.142 *10**-3 A\n"
- ]
- }
- ],
- "prompt_number": 51
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.14 Page no 557"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "C=100*10**-6\n",
- "R=40\n",
- "Ev=110\n",
- "f=60\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "Iv=Ev/(math.sqrt(R**2+(1/(2*math.pi*f*C)**2)))\n",
- "Iv1=math.sqrt(2)*Iv\n",
- "a=1/(2*math.pi*f*C*R)\n",
- "a1=math.atan(a)*180/3.14\n",
- "\n",
- "#Result\n",
- "print\"(a) Maximum current in the circuit is\", round(Iv1,2),\"A\"\n",
- "print\"(b) Phase lag between the current maximum and voltage maximum is\",round(a1,2),\"degree (e.m.f. lags behind the current)\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a) Maximum current in the circuit is 3.24 A\n",
- "(b) Phase lag between the current maximum and voltage maximum is 33.57 degree (e.m.f. lags behind the current)\n"
- ]
- }
- ],
- "prompt_number": 62
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.15 page no 557"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "C=100*10**-6\n",
- "R=50 #ohm\n",
- "L=0.5 #H\n",
- "Ev=110\n",
- "f=50\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "Z=math.sqrt(R**2+(2*math.pi*f*L-1/(2*math.pi*f*C))**2)\n",
- "I0=Ev/Z\n",
- "\n",
- "#Result\n",
- "print\"r.m.s value of current is\", round(I0,3),\"A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "r.m.s value of current is 0.816 A\n"
- ]
- }
- ],
- "prompt_number": 68
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.16 Page no 557"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "L=0.1 #H\n",
- "C=25*10**-6\n",
- "R=25.0\n",
- "e=314\n",
- "E0=310\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "f=e/(2*math.pi)\n",
- "Xl=2*math.pi*f*L\n",
- "Xc=1/(2*math.pi*f*C)\n",
- "A=Xc-Xl\n",
- "Z=math.sqrt(R**2+(Xc-Xl)**2)\n",
- "Ev=E0/math.sqrt(2)\n",
- "Iv=Ev/Z\n",
- "a1=(Xc-Xl)/R\n",
- "a2=math.atan(a1)*180/3.14\n",
- "a3=a2*math.pi/180.0\n",
- "V=Iv*Xc\n",
- "V1=Iv*Xl\n",
- "V2=Iv*R\n",
- "L=1/(((2*math.pi*f)**2)*C)\n",
- "\n",
- "#Result\n",
- "print\"(a) The frequency of the e.m.f is\",round(f,0),\"cycle s**-1\" \n",
- "print\"(b) The reactance of the circuit is\",round(A,0),\"ohm\"\n",
- "print\"(c) The impedance of the circuit is\",round(Z,1),\"ohm\"\n",
- "print\"(d) The current in the circuit is\",round(Iv,2),\"A\"\n",
- "print\"(e) The phase angle of the current is\",round(a3,3),\"rad\"\n",
- "print\"(f) The expression for the instantaneous value of the current is 3.125 cos(314t-1.316)\"\n",
- "print\"(g) Effective voltage across the capacitor is\",round(V,1),\"V\"\n",
- "print\" Effective Voltage across the inductor is\",round(V1,1),\"V\"\n",
- "print\"Effective voltage across the resistor is\",round(V2,2),\"V\"\n",
- "print\"(h) Value of inductance is\",round(L,3),\"H\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a) The frequency of the e.m.f is 50.0 cycle s**-1\n",
- "(b) The reactance of the circuit is 96.0 ohm\n",
- "(c) The impedance of the circuit is 99.2 ohm\n",
- "(d) The current in the circuit is 2.21 A\n",
- "(e) The phase angle of the current is 1.317 rad\n",
- "(f) The expression for the instantaneous value of the current is 3.125 cos(314t-1.316)\n",
- "(g) Effective voltage across the capacitor is 281.5 V\n",
- " Effective Voltage across the inductor is 69.4 V\n",
- "Effective voltage across the resistor is 55.25 V\n",
- "(h) Value of inductance is 0.406 H\n"
- ]
- }
- ],
- "prompt_number": 113
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.17 Page no 558"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "f=50 #Hz\n",
- "L=101.5*10**-3\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "C=1/((2*math.pi*f)**2*L)\n",
- "\n",
- "#Result\n",
- "print\"Capacitance of the capacitor is\",round(C*10**6,0),\"micro F\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Capacitance of the capacitor is 100.0 micro F\n"
- ]
- }
- ],
- "prompt_number": 117
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.18 Page no 558"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "Ev=230 #V\n",
- "L=5 #H\n",
- "C=80*10**-6\n",
- "R=40.0\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "f=1/(math.pi*2*math.sqrt(L*C))\n",
- "E0=math.sqrt(2)*Ev\n",
- "A=E0/R\n",
- "\n",
- "#Result\n",
- "print\"(i) Angular frequency is\", round(f,2),\"Hz\"\n",
- "print\"(ii) Impedence of circuit is\",R,\"ohm\"\n",
- "print\"(iii) Amplitude of the current is\",round(A,2)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(i) Angular frequency is 7.96 Hz\n",
- "(ii) Impedence of circuit is 40.0 ohm\n",
- "(iii) Amplitude of the current is 8.13\n"
- ]
- }
- ],
- "prompt_number": 126
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.19 Page no 558"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "C=2*10**-6 #F\n",
- "R=100 #ohm\n",
- "L=8 #H\n",
- "E=200 #V\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "F=1/((2*math.pi*math.sqrt(L*C)))\n",
- "D=2*math.pi*F*L\n",
- "L1=E/R\n",
- "\n",
- "#Result\n",
- "print\"The resonant frequency is\",round(F,2),\"Hz\"\n",
- "print\"(i) The inductive and capacitive reactances of the circuit is\",D,\"ohm\"\n",
- "print\"(ii) Total impedance of the circuit is 100\",\"ohm\"\n",
- "print\"(iii) Peak value of current is\",L1,\"A\"\n",
- "print\"(iv) The voltages across inductor and resistor differ in phase by\",\"90\"\n",
- "print\"(v) The voltages across inductor and capacitor differ in phase by\",\"180\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The resonant frequency is 39.79 Hz\n",
- "(i) The inductive and capacitive reactances of the circuit is 2000.0 ohm\n",
- "(ii) Total impedance of the circuit is 100 ohm\n",
- "(iii) Peak value of current is 2 A\n",
- "(iv) The voltages across inductor and resistor differ in phase by 90\n",
- "(v) The voltages across inductor and capacitor differ in phase by 180\n"
- ]
- }
- ],
- "prompt_number": 144
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.20 Page no 558"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "Ev=220 #V\n",
- "f=50 #Hz\n",
- "R=100.0 #ohm\n",
- "Vr=65 #V\n",
- "Vc=415 #V\n",
- "Vl=204 #V\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",
- "C1=1/(4*math.pi**2*f**2*L)\n",
- "\n",
- "#Result\n",
- "print\"(i) The current in the circuit is\",Iv,\"A\"\n",
- "print\"(ii) The value of the inductor is\",round(L,0),\"H\"\n",
- "print\"(iii) The value of the capacitor C is\",round(C*10**6,1),\"micro F\"\n",
- "print\"(iv) The value of C required to produce resonance is\",round(C1*10**6,1),\"micro F\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(i) The current in the circuit is 0.65 A\n",
- "(ii) The value of the inductor is 1.0 H\n",
- "(iii) The value of the capacitor C is 5.0 micro F\n",
- "(iv) The value of C required to produce resonance is 10.1 micro F\n"
- ]
- }
- ],
- "prompt_number": 189
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.21 Page no 559"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "L=3 #H\n",
- "C=27*10**-6\n",
- "R=7.4 #ohm\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "w0=1/(math.sqrt(L*C))\n",
- "Q=1/R*(math.sqrt(L/C))\n",
- "\n",
- "#Result\n",
- "print\"Resonant frequency is\", round(w0,1),\"rad/s\"\n",
- "print\"Q factor is\",round(Q,2)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Resonant frequency is 111.1 rad/s\n",
- "Q factor is 45.05\n"
- ]
- }
- ],
- "prompt_number": 196
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.22 Page no 559"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "R=30 #ohm\n",
- "Xl=40\n",
- "E0=220\n",
- "I0=1 \n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "Z=math.sqrt(R**2+Xl**2)\n",
- "a=R/Z\n",
- "Pav=E0*I0*a/(math.sqrt(2)*math.sqrt(2))\n",
- "\n",
- "#Result\n",
- "print\"Power consumed in the circuit is\", Pav,\"Watt\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Power consumed in the circuit is 66.0 Watt\n"
- ]
- }
- ],
- "prompt_number": 201
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.23 Page no 559"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "Ev=100\n",
- "f=50 #Hz\n",
- "C=10*10**-6\n",
- "R=100\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "Xc=1/(2*math.pi*f*C)\n",
- "Iv=Ev/Xc\n",
- "Pav=Ev*Iv*Ev/(math.sqrt(Ev**2+Xc**2))\n",
- "\n",
- "#Result\n",
- "print\"(a) The reactance of the capacitor is\",round(Xc,2),\"ohm\"\n",
- "print\"(b) Current flowing is\",round(Iv,3),\"A\"\n",
- "print\"(c) Average power supplied is\",round(Pav,2),\"W\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a) The reactance of the capacitor is 318.31 ohm\n",
- "(b) Current flowing is 0.314 A\n",
- "(c) Average power supplied is 9.42 W\n"
- ]
- }
- ],
- "prompt_number": 214
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.24 Page no 559"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "V=60.0\n",
- "P=10\n",
- "Ev=100\n",
- "f=60 #Hz\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "I=P/V\n",
- "R=f/I\n",
- "Z=Ev/I\n",
- "L=math.sqrt(Z**2-R**2)/(2*math.pi*f)\n",
- "R1=Z-R\n",
- "\n",
- "#Result\n",
- "print\"(i) The inductance is\", round(L,3),\"henry\"\n",
- "print\"(ii) Value of resistance is\",R1,\"ohm\"\n",
- "print\"(iii) If resistance is used in the place of inductance, the electrical energy is wasted.\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(i) The inductance is 1.273 henry\n",
- "(ii) Value of resistance is 240.0 ohm\n",
- "(iii) If resistance is used in the place of inductance, the electrical energy is wasted.\n"
- ]
- }
- ],
- "prompt_number": 224
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.25 Page no 559"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "V=50.0 #V\n",
- "P=20 #watt\n",
- "Ev=250 #V\n",
- "f=50\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "I=P/V\n",
- "R=V/I\n",
- "Z=Ev/I\n",
- "C=1/(2*math.pi*f*Ev*math.sqrt(6))\n",
- "\n",
- "#Result\n",
- "print\"Value of capacitance required is\", round(C*10**6,3)*10**-6,\"F\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Value of capacitance required is 5.198e-06 F\n"
- ]
- }
- ],
- "prompt_number": 233
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.26 Page no 560"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "Ev=200\n",
- "f=50 #H\n",
- "R=50\n",
- "L=0.3\n",
- "C=40*10**-6\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "Xl=2*math.pi*f*L\n",
- "Xc=1/(2*math.pi*f*C)\n",
- "Z=math.sqrt(R**2+(Xl-Xc)**2)\n",
- "Iv=Ev/Z\n",
- "a=R/Z\n",
- "Pav=Ev*Iv*a\n",
- "\n",
- "#Result\n",
- "print\"Impedence in the circuit is\", round(Z,2),\"ohm\"\n",
- "print\"Power in the circuit is\",round(Pav,1),\"watt\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Impedence in the circuit is 52.11 ohm\n",
- "Power in the circuit is 736.6 watt\n"
- ]
- }
- ],
- "prompt_number": 244
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.27 Page no 560"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "Ev=200 #v\n",
- "L=5 #H\n",
- "C=80\n",
- "R=40.0\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "W=10**3/(math.sqrt(L*C))\n",
- "Iv=Ev/R\n",
- "I0=math.sqrt(2)*Iv\n",
- "Pav=Ev*L*math.cos(0*3.14/180.0)\n",
- "\n",
- "print\"(a) Angular frequency is\",W,\"rad s**-1\"\n",
- "print\"(b) The current amplitude is\",round(I0,2),\"A\" \n",
- "print\"(c) The power dissipation in the circuit is\",Pav,\"Watt\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a) Angular frequency is 50.0 rad s**-1\n",
- "(b) The current amplitude is 7.07 A\n",
- "(c) The power dissipation in the circuit is 1000.0 Watt\n"
- ]
- }
- ],
- "prompt_number": 272
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.28 Page no 560"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "L=5.0 #H\n",
- "C=80*10**-6 #F\n",
- "R=40.0 #ohm\n",
- "Ev=230 #V\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "D=1/math.sqrt(L*C)\n",
- "Iv=Ev/R\n",
- "I0=math.sqrt(2)*Iv\n",
- "S=Iv*R\n",
- "S1=Iv*D*L\n",
- "S2=Iv/(1/D*C)\n",
- "S3=Iv/(D*L-1/D*C)\n",
- "\n",
- "#Result\n",
- "print\"(a) The resonant angular frequency is\",D,\"rad s**-1\"\n",
- "print\"(b) The impedance of the circuit and the amplitude of curremt is\",round(I0,2),\"A\"\n",
- "print\"(c) The R.M.S. potential drop across LC is\",round(S3,0)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a) The resonant angular frequency is 50.0 rad s**-1\n",
- "(b) The impedance of the circuit and the amplitude of curremt is 8.13 A\n",
- "(c) The R.M.S. potential drop across LC is 0.0\n"
- ]
- }
- ],
- "prompt_number": 282
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.29 Page no 560"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "L=0.12 #H\n",
- "C=480*10**-9 #F\n",
- "R=23 #ohm\n",
- "Ev=230 #V\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "D=1/(math.sqrt(L*C))\n",
- "Iv=Ev/(math.sqrt(R**2+(D*L-1/D*C))**2)\n",
- "I0=(math.sqrt(2)*Ev)/(math.sqrt(R**2+(D*L-1/D*C))**2)\n",
- "I1=(math.sqrt(2)*Ev)/R\n",
- "EvIv=Ev*(I1/math.sqrt(2))\n",
- "\n",
- "#Result\n",
- "print\"(a) The source frequency is\",round(D,1),\"rad s**-1\"\n",
- "print\" The maximum value is\",round(I1,2),\"A\"\n",
- "print\"(b) Average power will also be maximum at resonant frequency is\",round(D,1),\"rad s**-1\" \n",
- "print\" The value of this maximum power is\",EvIv,\"Watt\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a) The source frequency is 4166.7 rad s**-1\n",
- " The maximum value is 14.14 A\n",
- "(b) Average power will also be maximum at resonant frequency is 4166.7 rad s**-1\n",
- " The value of this maximum power is 2300.0 Watt\n"
- ]
- }
- ],
- "prompt_number": 298
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 18.30 Page no 561"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "L=80*10**-3 #H\n",
- "C=60*10**-6 #F\n",
- "Ev=230 #V\n",
- "f=50 #Hz\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "Iv=-Ev/((2*math.pi*f*L)-(1/(2*math.pi*f*C)))\n",
- "I0=math.sqrt(2)*Iv\n",
- "Iv1=Iv*2*math.pi*f*L\n",
- "Iv2=Iv*(1/(2*math.pi*f*C))\n",
- "Pav=math.cos(90*3.14/180.0)*Ev\n",
- "Pav1=math.cos(-90*3.14/180.0)*Ev\n",
- "\n",
- "#Result\n",
- "print\"(a) The current amplitude is\",round(Iv,2),\"A\",\"and r.m.s. value is\",round(I0,2),\"A\"\n",
- "print\"(b) The r.m.s. value of potential drops across L is\",round(Iv1,1),\"V\",\"and across C is\",round(Iv2,1),\"W\"\n",
- "print\"(c) The average power transferred to the inductor is\",round(Pav,0)\n",
- "print\"(d) The average power transferred to the capacitor is\",round(Pav1,0)\n",
- "print\"(e) Total average power absorbed is zero\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a) The current amplitude is 8.24 A and r.m.s. value is 11.65 A\n",
- "(b) The r.m.s. value of potential drops across L is 207.0 V and across C is 437.0 W\n",
- "(c) The average power transferred to the inductor is 0.0\n",
- "(d) The average power transferred to the capacitor is 0.0\n",
- "(e) Total average power absorbed is zero\n"
- ]
- }
- ],
- "prompt_number": 357
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/modern_physics_by_Satish_K._Gupta/chap19.ipynb b/modern_physics_by_Satish_K._Gupta/chap19.ipynb deleted file mode 100755 index 6e89d5a2..00000000 --- a/modern_physics_by_Satish_K._Gupta/chap19.ipynb +++ /dev/null @@ -1,640 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:c81057d61d55f846d72bc0a6263de0452d74cd37dd94b021ce01d87a17a0c974"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 19 Electrical machines and devices"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 19.1 Page no 582"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "Ep=2300.0 #v\n",
- "Np=4000\n",
- "Es=230 #v\n",
- "\n",
- "#Calculation\n",
- "Ns=Np*(Es/Ep)\n",
- "\n",
- "#Result\n",
- "print\"Number of turns is\", Ns"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Number of turns is 400.0\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 19.2 Page no 582"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "Ep=220.0 #v\n",
- "Es=110\n",
- "E=550 #watt\n",
- "\n",
- "#Calculation\n",
- "Ip=E/Ep\n",
- "\n",
- "#Result\n",
- "print\"Current drawn is\",Ip,\"A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Current drawn is 2.5 A\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 19.3 Page no 582"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Gien\n",
- "Ep=220.0 #v\n",
- "Es=22 \n",
- "Z=220.0 #ohm\n",
- "\n",
- "#Calculation\n",
- "Is=Es/Z\n",
- "Ip=(Es/Ep)*Is\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": 6
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 19.4 Page no 582"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "Np=200.0\n",
- "Ns=1000\n",
- "Ep=200.0 #v\n",
- "E=10**4 #W\n",
- "\n",
- "#Calculation\n",
- "Es=(Ns/Np)*Ep\n",
- "Ip=E/Ep\n",
- "\n",
- "#Result\n",
- "print\"(a) Output voltage is\",Es,\"V\"\n",
- "print\"(b) Current is\",Ip,\"A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a) Output voltage is 1000.0 V\n",
- "(b) Current is 50.0 A\n"
- ]
- }
- ],
- "prompt_number": 12
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 19.5 Page no 582"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "ep=120\n",
- "Ip=1.85 #A\n",
- "Is=150*10**-3\n",
- "n=0.95\n",
- "\n",
- "#Calculation\n",
- "es=n*ep*Ip/Is\n",
- "\n",
- "#result\n",
- "print\"Voltage across the secondary is\",es,\"V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Voltage across the secondary is 1406.0 V\n"
- ]
- }
- ],
- "prompt_number": 14
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 19.6 Page no 582"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "K=100\n",
- "Np=100\n",
- "e=1100 #watt\n",
- "ep=220.0\n",
- "\n",
- "#Calculation\n",
- "Ns=K*Np\n",
- "ip=e/ep\n",
- "es=K*ep\n",
- "Is=ip*ep/es\n",
- "\n",
- "#Result\n",
- "print\"(i) Number of turns is\",Ns\n",
- "print\"(ii) Current is\",ip,\"A\"\n",
- "print\"(iii) Voltage across the secondary is\",es,\"V\"\n",
- "print\"(iv) Current in the secondaryis\",Is,\"A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(i) Number of turns is 10000\n",
- "(ii) Current is 5.0 A\n",
- "(iii) Voltage across the secondary is 22000.0 V\n",
- "(iv) Current in the secondaryis 0.05 A\n"
- ]
- }
- ],
- "prompt_number": 23
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 19.7 Page no 583"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "e=11000 #W\n",
- "es=220.0 #V\n",
- "es1=22000.0\n",
- "\n",
- "#Calculation\n",
- "Is=e/es\n",
- "V=Is**2\n",
- "Is1=e/es1\n",
- "P=Is1**2\n",
- "\n",
- "#Result\n",
- "print\"(i) Electrical energy dissipated when power is transmitted at es=220 V is\",V,\"R watt\"\n",
- "print\"(ii)Electrical energy dissipated when power is transmitted at es=22000 V is\",P,\"R watt\"\n",
- "print\"Transmission should be done at 22000 V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(i) Electrical energy dissipated when power is transmitted at es=220 V is 2500.0 R watt\n",
- "(ii)Electrical energy dissipated when power is transmitted at es=22000 V is 0.25 R watt\n",
- "Transmission should be done at 22000 V\n"
- ]
- }
- ],
- "prompt_number": 28
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 19.8 Page no 583"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "B=0.1 #T\n",
- "n=2000\n",
- "A=0.05 #m**2\n",
- "w=2100 #r.p.m.\n",
- "\n",
- "#calculation\n",
- "import math\n",
- "A1=2*math.pi*w/60.0\n",
- "e0=w*B*A*A1\n",
- "e=e0*math.sin(60*3.14/180.0)\n",
- "\n",
- "#Result\n",
- "print\"Maximum e.m.f is\", round(e0,0),\"V\"\n",
- "print\"Instantaneous e.m.f is\",round(e,0),\"V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Maximum e.m.f is 2309.0 V\n",
- "Instantaneous e.m.f is 1999.0 V\n"
- ]
- }
- ],
- "prompt_number": 50
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 19.9 Page no 583"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "n=100\n",
- "A=3 #m**2\n",
- "w=60 #rad/s\n",
- "B=0.04\n",
- "R=500.0\n",
- "\n",
- "#Calculation\n",
- "e0=n*B*A*w\n",
- "I0=e0/R\n",
- "P=e0*I0\n",
- "\n",
- "#Result\n",
- "print\"(a) Maximum current is\",I0,\"A\"\n",
- "print\"(b) Maximum power is\",P,\"W\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a) Maximum current is 1.44 A\n",
- "(b) Maximum power is 1036.8 W\n"
- ]
- }
- ],
- "prompt_number": 56
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 19.10 Page no 583"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "n=50\n",
- "A=2.5 #m**2\n",
- "w=60 #rad/s\n",
- "B=0.30\n",
- "R=500.0\n",
- "\n",
- "#Calculation\n",
- "e0=n*B*A*w\n",
- "I0=e0/R\n",
- "\n",
- "#Result\n",
- "print\"(a) The maximum current is\",I0,\"A\"\n",
- "print\"(b) Current is zero when coil is vertical and flux is maximum.On the other hand current is maximum when coil is horizontal,flux is minimum\"\n",
- "print\"(c) Yes,there should be relative motion between the magnetic field and coil\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a) The maximum current is 4.5 A\n",
- "(b) Current is zero when coil is vertical and flux is maximum.On the other hand current is maximum when coil is horizontal,flux is minimum\n",
- "(c) Yes,there should be relative motion between the magnetic field and coil\n"
- ]
- }
- ],
- "prompt_number": 60
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 19.11 Page no 583"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "E=220\n",
- "e=200.0\n",
- "P=5000\n",
- "\n",
- "#Calculation\n",
- "I=P/e\n",
- "R=(E-e)/I\n",
- "\n",
- "#Result\n",
- "print\"Armature current is\",I,\"A\"\n",
- "print\"Motor resistance is\",R,\"ohm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Armature current is 25.0 A\n",
- "Motor resistance is 0.8 ohm\n"
- ]
- }
- ],
- "prompt_number": 65
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 19.12 Page no 583"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "R=8.5\n",
- "E=200\n",
- "I=5\n",
- "\n",
- "#Calculation\n",
- "e=E-(I*R)\n",
- "Pi=E*I\n",
- "Po=E*I-(I**2*R)\n",
- "n=Po/Pi\n",
- "\n",
- "#Result\n",
- "print\"Back e.m.f of the motor is\", e,\"V\"\n",
- "print\"Power input is\",Pi,\"W\"\n",
- "print\"Power output is\",Po,\"W\"\n",
- "print\"Efficiencyis\",n*10**2,\"%\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Back e.m.f of the motor is 157.5 V\n",
- "Power input is 1000 W\n",
- "Power output is 787.5 W\n",
- "Efficiencyis 78.75 %\n"
- ]
- }
- ],
- "prompt_number": 78
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 19.13 Page no 584"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "h=300\n",
- "g=9.8 #m/s**2\n",
- "a=1000\n",
- "A=100.0\n",
- "n=60\n",
- "\n",
- "#Calculation\n",
- "m=A*a\n",
- "E=m*g*h\n",
- "P=E*n/A\n",
- "\n",
- "#Result\n",
- "print\"Electric power is\", P*10**-6,\"MW\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Electric power is 176.4 MW\n"
- ]
- }
- ],
- "prompt_number": 84
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 19.14 Page no 584"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "L=30 #km\n",
- "R=15 #ohm\n",
- "V=4000.0 #V\n",
- "P=8*10**5 #W\n",
- "\n",
- "#Calculation\n",
- "Is=P/V\n",
- "P1=Is**2*R\n",
- "P2=P+P1\n",
- "V1=R*Is\n",
- "A=440-(V+V1)\n",
- "\n",
- "#Result\n",
- "print\"(a) Line power loss is\",P1*10**-3,\"KW\"\n",
- "print\"(b) Supplied power is\",P2*10**-3,\"KW\"\n",
- "print\"(c) Step up transformer at the plane is\",A"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a) Line power loss is 600.0 KW\n",
- "(b) Supplied power is 1400.0 KW\n",
- "(c) Step up transformer at the plane is -6560.0\n"
- ]
- }
- ],
- "prompt_number": 94
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 19.15 Page no 584"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "r=8*10**-2\n",
- "n=20\n",
- "w=50 #rad/s\n",
- "B=3*10**-2 #T\n",
- "A=64*math.pi*10**-4 #m**2\n",
- "Eav=0\n",
- "R=10 #ohm\n",
- "\n",
- "#Calculation\n",
- "math.pi\n",
- "e0=n*B*A*w\n",
- "Pav=e0**2/(2*R)\n",
- "\n",
- "#Result\n",
- "print\" Maximum e.m.f. is\",round(e0,4),\"V\"\n",
- "print\" Average e.m.f. is zero\"\n",
- "print\" Dissipated power is\",round(Pav,4),\"W\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- " Maximum e.m.f. is 0.6032 V\n",
- " Average e.m.f. is zero\n",
- " Dissipated power is 0.0182 W\n"
- ]
- }
- ],
- "prompt_number": 103
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/modern_physics_by_Satish_K._Gupta/chap2.ipynb b/modern_physics_by_Satish_K._Gupta/chap2.ipynb deleted file mode 100755 index efbea0a7..00000000 --- a/modern_physics_by_Satish_K._Gupta/chap2.ipynb +++ /dev/null @@ -1,534 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:1e2d45d8326c66f64e42e0f8f482f41d5aabf3a1eabc7e50f07a0c18121fe999"
- },
- "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 39"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "m=10**-7 #Kg\n",
- "q=1.6*10**-19 #C\n",
- "g=9.8\n",
- "\n",
- "#Calculation\n",
- "E=(m*g)/q\n",
- "\n",
- "#Result\n",
- "print\"Electric field strength required is\", E"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Electric field strenght required is 6.125e+12\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 2.2 Page no 39"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "m=10.0**-3 #Kg\n",
- "q=5*10**-6 #C\n",
- "u=20 #m/s\n",
- "E=2*10**5 #N/C\n",
- "\n",
- "#Calculation\n",
- "F=q*E\n",
- "a=-F/m\n",
- "S=-u**2/(2.0*a)\n",
- "\n",
- "#Result\n",
- "print\"Distance travelled is\", S,\"m\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Distance travelled is 0.2 m\n"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 2.3 Page no 39"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "E=2.55*10**4 #V/m\n",
- "a=1.26*10**3 #Kg/m**4\n",
- "g=9.81 #m/s**2\n",
- "e=1.6*10**-19\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "q=12*e\n",
- "Fe=q*E\n",
- "r=((3*Fe)/(4.0*math.pi*a*g))**0.333\n",
- "\n",
- "#Result\n",
- "print\"The radius of the drop is\", round(r*10**4,5)*10**3,\"*10**-4 mm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The radius of the drop is 9.95 *10**-4 mm\n"
- ]
- }
- ],
- "prompt_number": 28
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 2.4 Page no 39"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "m=80.0*10**-6 #Kg\n",
- "q=2*10**-8 #C\n",
- "E=20000 #V/m\n",
- "g=9.8\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "a=(q*E)/(m*g)\n",
- "a1=math.atan(a)*180/3.14\n",
- "T=(q*E)/(math.sin(a1*3.14/180.0))\n",
- "\n",
- "#Result\n",
- "print\"Tension in the thread of the pendulum is\", round(T*10**4,1),\"*10**-4 N\"\n",
- "print\"Angle it make with the vertical is\",round(a1,2),\"Degree C\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Tension in the thread of the pendulum is 8.8 *10**-4 N\n",
- "Angle it make with the vertical is 27.04 Degree C\n"
- ]
- }
- ],
- "prompt_number": 41
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 2.5 Page no 40"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "q1=2.0*10**-7 #C\n",
- "q2=1.0*10**-7 #C\n",
- "r=10.0**-2 #m\n",
- "m=9*10**9\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "E1=(m*q1)/r**2\n",
- "E2=(m*q2)/r**2\n",
- "\n",
- "#Result\n",
- "print\"Magnitude of field E1 is\", E1*10**-7,\"*10**7 N/C\"\n",
- "print\"Magnitude of field E2 is\", E2*10**-6,\"*10**6 N/C\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Magnitude of field E1 is 1.8 *10**7 N/C\n",
- "Magnitude of field E2 is 9.0 *10**6 N/C\n"
- ]
- }
- ],
- "prompt_number": 50
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 2.6 Page no 40"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "q1=5.0*10**-19 #c\n",
- "q2=20*10**-19\n",
- "r=2.0 #m\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "x=-math.sqrt(q2/q1)\n",
- "x1=math.sqrt(q2/(q1*3))\n",
- "\n",
- "#Result\n",
- "print\"The point on the line joining is\", round(x,1),\"m and \",round(x1,1),\"m\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The point on the line joining is -2.0 m and 1.2 m\n"
- ]
- }
- ],
- "prompt_number": 60
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 2.7 Page no 40"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "r=0.2\n",
- "m=9*10**9\n",
- "qa=3*10**-6\n",
- "q=1.5*10**-9\n",
- "\n",
- "#Calculation\n",
- "a=r/2.0\n",
- "Ea=(m*qa)/a**2\n",
- "Eb=(m*qa)/a**2\n",
- "E=Ea+Eb\n",
- "F=q*E\n",
- "\n",
- "#Result\n",
- "print\"(a) Electric field at the mid point is\", E*10**-6,\"*10**6 N/C (along OB)\"\n",
- "print\"(b) Force experienced by the test charge is\",F*10**3,\"*10**-3 N (along OA)\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a) Electric field at the mid point is 5.4 *10**6 N/C (along OB)\n",
- "(b) Force experienced by the test charge is 8.1 *10**-3 N (along OA)\n"
- ]
- }
- ],
- "prompt_number": 71
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 2.8 Page no 40"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "qa=16*10**-6 #c\n",
- "qb=-9.0*10**-6 #C\n",
- "r=0.08 #m\n",
- "\n",
- "#Calculation\n",
- "x=-(qa/qb)*r\n",
- "\n",
- "#Result\n",
- "print\"Position of the point is\", round(x,2),\"m\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Position of the point is 0.14 m\n"
- ]
- }
- ],
- "prompt_number": 82
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 2.9 Page no 41"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "m=9*10**9\n",
- "qa=4*10**-6\n",
- "r=2.0*10**-2\n",
- "r1=10**-2\n",
- "qb=2*10**-6\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "Ea=(m*qa)/r**2\n",
- "Eb=(m*qb)/r1**2\n",
- "a=r1/r\n",
- "E=math.sqrt(Ea**2+Eb**2+(2*Ea*Eb*a))\n",
- "a11=Ea*(math.sin(60)*180/3.14)/(Ea+Eb*(math.cos(60)*180/3.14))\n",
- "\n",
- "#Result\n",
- "print\"Position is\", round(a11,2)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Position is 0.16\n"
- ]
- }
- ],
- "prompt_number": 33
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 2.10 Page no 41"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "q=500*10**-6\n",
- "r=0.1 #m\n",
- "d=0.2\n",
- "m=9*10**9\n",
- "\n",
- "#Calculation\n",
- "p=q*r\n",
- "a=r/2.0\n",
- "R=d+a\n",
- "E=(m*2*p*R)/((R**2-a**2)**2)\n",
- "\n",
- "#Result\n",
- "print\"Electric dipole moment of the dipole is\", p,\"Cm\"\n",
- "print\"Electric field due to dipole is\",E*10**-7,\"*10**7 N/C\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Electric dipole moment of the dipole is 5e-05 Cm\n",
- "Electric field due to dipole is 6.25 *10**7 N/C\n"
- ]
- }
- ],
- "prompt_number": 21
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 2.11 Page no 41"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "q=0.2*10**-12 #C\n",
- "w=10**-8 #m\n",
- "r=0.1 #m\n",
- "e=9*10**9\n",
- "\n",
- "#Calculation\n",
- "p=q*w\n",
- "E=(e*2*p)/r**3\n",
- "\n",
- "#Result\n",
- "print\"The electric field at an axical point at a distance of 10cm from there mid point is\", E*10**8,\"*10**-8 N/C\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The electric field at an axical point at a distance of 10cm from there mid point is 3.6 *10**-8 N/C\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 2.13 Page no 42"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "E=10.0**4\n",
- "t=9*10**-26 #Nm\n",
- "\n",
- "#Calculation\n",
- "import math \n",
- "p=t/(E*math.sin(30*3.14/180.0))\n",
- "\n",
- "#Result\n",
- "print\"Dipole moment of the dipole is\", round(p*10**29,1),\"*10**-19 Cm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Dipole moment of the dipole is 1.8 *10**-19 Cm\n"
- ]
- }
- ],
- "prompt_number": 10
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 2.17 Page no 43"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "p=-10**-7 #Cm\n",
- "i=10**5\n",
- "\n",
- "#Calculation\n",
- "F=p*i\n",
- "\n",
- "#Result\n",
- "print\"Force is given by\", F,\"N\"\n",
- "print\"Torque is zero\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Force is given by -0.01 N\n",
- "Torque is zero\n"
- ]
- }
- ],
- "prompt_number": 15
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/modern_physics_by_Satish_K._Gupta/chap20.ipynb b/modern_physics_by_Satish_K._Gupta/chap20.ipynb deleted file mode 100755 index b99b0934..00000000 --- a/modern_physics_by_Satish_K._Gupta/chap20.ipynb +++ /dev/null @@ -1,255 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:17722fc50855830105e8bbc1d5e7f4e9c2450186689d4bcffcae55541cf58a10"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 20 Electromagnetic Waves"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 20.1 Page no 618"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "L=5.5*10**-7 #m\n",
- "c=3*10**8 #m s**-1\n",
- "u=1.5\n",
- "\n",
- "#Calculation\n",
- "V=c/L\n",
- "T=L/c\n",
- "v=c/u\n",
- "L1=v*T\n",
- "\n",
- "#Result\n",
- "print\"(a) The frequency is\",round(V*10**-14,3),\"10**8\",\"MHz\",\"and time period is\",round(T*10**15,1),\"10**-19\",\"micro s\"\n",
- "print\"(b) The wavelenght in glass is\",round(L1*10**7,1),\"10**-7\",\"m\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a) The frequency is 5.455 10**8 MHz and time period is 1.8 10**-19 micro s\n",
- "(b) The wavelenght in glass is 3.7 10**-7 m\n"
- ]
- }
- ],
- "prompt_number": 20
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 20.2 Page no 619"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "V=2*10**8 #m s**-1\n",
- "Ur=1.0\n",
- "C=3*10**8\n",
- "\n",
- "#Calculation\n",
- "Er=(C**2)/(V**2*Ur)\n",
- "\n",
- "#Result \n",
- "print\" The relative permittivity is\",Er"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- " The relative permittivity is 2.25\n"
- ]
- }
- ],
- "prompt_number": 24
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 20.3 Page no 619"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "C=3*10**8\n",
- "V=10**8\n",
- "E0=0.5\n",
- "\n",
- "#Calculation\n",
- "L=C/V\n",
- "Bz=E0/C\n",
- "\n",
- "#Result\n",
- "print\"(a) There is the positive direction of x-axis of propagation of electromagnetic waves\"\n",
- "print\"(b) The wavelengh of the wave is\",L,\"m\"\n",
- "print\"(c) The component of associated magnetic field is\",round(Bz*10**9,0),\"cos(2*math.pi*10**8(t-x/c))\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a) There is the positive direction of x-axis of propagation of electromagnetic waves\n",
- "(b) The wavelengh of the wave is 3 m\n",
- "(c) The component of associated magnetic field is 2.0 cos(2*math.pi*10**8(t-x/c))\n"
- ]
- }
- ],
- "prompt_number": 35
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 20.4 Page no 619"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "L=6*10**-3 #m\n",
- "E0=33 #V m**-1\n",
- "C=3.0*10**8\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "W=(2*math.pi*C)/(L)\n",
- "B=E0/C\n",
- "\n",
- "#Result\n",
- "print\"Magnetic field is\",B,\"sin*math.pi*10**11(t-x/c)\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Magnetic field is 1.1e-07 sin*math.pi*10**11(t-x/c)\n"
- ]
- }
- ],
- "prompt_number": 41
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 20.5 Page no 619"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "I=8\n",
- "E0=8.85*10**-12\n",
- "C=3*10**8\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "e0=math.sqrt((2*I)/(E0*C))\n",
- "\n",
- "#Result\n",
- "print\"The amplitude of the electric field is\",round(e0,1),\"N C**-1\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The amplitude of the electric field is 77.6 N C**-1\n"
- ]
- }
- ],
- "prompt_number": 46
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 20.6 Page no 619"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "L=0.024 #m\n",
- "k=5.9*10**-9\n",
- "c=3*10**8\n",
- "\n",
- "#Calculation\n",
- "V=L/k\n",
- "t=L/c\n",
- "\n",
- "#Result\n",
- "print\"(i) The number of oscillation is\",round(V*10**-6,3),\"10**6\",\"Hz\"\n",
- "print\"(ii) The coherence time is\",t,\"s\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(i) The number of oscillation is 4.068 10**6 Hz\n",
- "(ii) The coherence time is 8e-11 s\n"
- ]
- }
- ],
- "prompt_number": 65
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/modern_physics_by_Satish_K._Gupta/chap21.ipynb b/modern_physics_by_Satish_K._Gupta/chap21.ipynb deleted file mode 100755 index 54cafbe9..00000000 --- a/modern_physics_by_Satish_K._Gupta/chap21.ipynb +++ /dev/null @@ -1,1809 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:07c4b928a455230c9db333bb0062ec9efe3c0f7899f2ff1857d420b5212b1285"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 21 Refraction Of Light"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 21.1 Page no 650"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "U=1.5\n",
- "c=3.0*10**8 #m s**-1\n",
- "\n",
- "#Calculation\n",
- "v=c/U\n",
- "\n",
- "#Result\n",
- "print\" Speed of light in glass is\",v*10**-8,\"10**8\",\"m s**-1\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- " Speed of light in glass is 2.0 10**8 m s**-1\n"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 21.2 Page no 650"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "u=1.33\n",
- "Lu=589*10**-9 #m\n",
- "c=3*10**8\n",
- "\n",
- "#Calculation\n",
- "v=c/Lu\n",
- "v1=c/u\n",
- "Lw=v1/v\n",
- "\n",
- "#Result\n",
- "print\"(a) Wavelenght of light after reflection is\",Lu*10**9,\"10**-9\"\n",
- "print\" Velocity of light after reflection is\",c*10**-8,\"10**8\",\"m s**-1\"\n",
- "print\" Frqequency of light after reflection is\",round(v*10**-14,4),\"10**14\",\"Hz\"\n",
- "print\"(b) Frequency of light after refraction is\",round(v*10**-14,4),\"10**14\",\"Hz\"\n",
- "print\" Velocity of light after refraction is\",round(v1*10**-8,3),\"10**8\",\"m s**-1\"\n",
- "print\" Wavelength of light after refraction is\",round(Lw*10**9,2),\"nm\" "
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a) Wavelenght of light after reflection is 589.0 10**-9\n",
- " Velocity of light after reflection is 3.0 10**8 m s**-1\n",
- " Frqequency of light after reflection is 5.0934 10**14 Hz\n",
- "(b) Frequency of light after refraction is 5.0934 10**14 Hz\n",
- " Velocity of light after refraction is 2.256 10**8 m s**-1\n",
- " Wavelength of light after refraction is 442.86 nm\n"
- ]
- }
- ],
- "prompt_number": 35
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 21.3 Page no 651"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "i=60\n",
- "u=1.5\n",
- "t=0.1 #m\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "r=(math.sin(60*3.14/180.0))/u\n",
- "a=math.asin(r)*180/3.14\n",
- "d=t/math.cos(r*3.14/180.0)*(math.sin(24*3.14/180.0))\n",
- "\n",
- "#Result\n",
- "print\"The lateral shift produced is\",round(d,4),\"m\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The lateral shift produced is 0.0407 m\n"
- ]
- }
- ],
- "prompt_number": 48
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 21.4 Page no 651"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "A=4/3.0\n",
- "A1=3/2.0\n",
- "I=30\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "S=A1/A\n",
- "R=math.sin(30*3.16/180.0)/S\n",
- "A=math.asin(R)*180/3.14\n",
- "\n",
- "#Result\n",
- "print\"The angle of refraction is\",round(A,0)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The angle of refraction is 27.0\n"
- ]
- }
- ],
- "prompt_number": 59
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 21.5 Page no 651"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "t=15\n",
- "u=1.5\n",
- "\n",
- "#Calculation\n",
- "d=t*(1-(1/u))\n",
- "\n",
- "#Result\n",
- "print\" Real thickness of glass slab is\",d,\"cm\"\n",
- "print\" The answer does not depend upon the location of the slab\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- " Real thickness of glass slab is 5.0 cm\n",
- " The answer does not depend upon the location of the slab\n"
- ]
- }
- ],
- "prompt_number": 67
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 21.6 Page no 651"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "c=3*10**8 #m s**-1\n",
- "v=2.0*10**8 #m s**-1\n",
- "t=6.0 #cm\n",
- "\n",
- "#Calculation\n",
- "u=c/v\n",
- "d=t*(1-(1/u))\n",
- "\n",
- "#Result\n",
- "print\"An ink dot appear to be rasied\",d,\"cm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "An ink dot appear to be rasied 2.0 cm\n"
- ]
- }
- ],
- "prompt_number": 73
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 21.7 Page no 651"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "d=12.5 #cm\n",
- "a=9.4 #cm\n",
- "u=1.63\n",
- "\n",
- "#Calculation\n",
- "S=d/a\n",
- "S1=d/u\n",
- "S3=a-S1\n",
- "\n",
- "#Result\n",
- "print\"The refractive index of water is\",round(S,2)\n",
- "print\"The refractive index of liquid is\",round(S1,2),\"cm\"\n",
- "print\"Distance is\",round(S3,2),\"cm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The refractive index of water is 1.33\n",
- "The refractive index of liquid is 7.67 cm\n",
- "Distance is 1.73 cm\n"
- ]
- }
- ],
- "prompt_number": 84
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 21.8 Page no 651"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "A=4/3.0\n",
- "d=0.015 #m\n",
- "\n",
- "#Calculatiom\n",
- "t=d/(1-(1/A))\n",
- "\n",
- "#Result\n",
- "print\"The height upto which water must be poured into the beaker is\",t,\"m\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The height upto which water must be poured into the beaker is 0.06 m\n"
- ]
- }
- ],
- "prompt_number": 91
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 21.9 Page no 652"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "A=3/2.0\n",
- "A1=4/3.0\n",
- "t1=6\n",
- "t2=4\n",
- "\n",
- "#Calculation\n",
- "d1=t1*(1-(1/A))\n",
- "d2=t2*(1-(1/A1))\n",
- "d3=d1+d2\n",
- "\n",
- "#Result\n",
- "print\"The apparent position of an object is\",d3,\"cm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The apparent position of an object is 3.0 cm\n"
- ]
- }
- ],
- "prompt_number": 102
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 21.10 Page no 652"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "S=1.5\n",
- "W=1.33\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "A=S/W\n",
- "C=1/A\n",
- "Q=math.asin(C)*180/3.14\n",
- "\n",
- "#Result\n",
- "print\"The critical angle for a glass water interface is\",round(Q,0),\"degree\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The critical angle for a glass water interface is 62.0 degree\n"
- ]
- }
- ],
- "prompt_number": 124
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 21.11 Page no 652"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "I=40 #degree\n",
- "A=15 #degree\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "r=I-A\n",
- "u=(math.sin(40*3.14/180.0))/(math.sin(25*3.14/180.0))\n",
- "c=1/u\n",
- "A=math.asin(c)*180/3.14\n",
- "\n",
- "#Result\n",
- "print\"The critical angle is\",round(A,1),\"degree\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The critical angle is 41.1 degree\n"
- ]
- }
- ],
- "prompt_number": 123
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 21.12 Page no 652"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "u=1.5\n",
- "h=20\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "C=1/u\n",
- "A=math.asin(C)*180/3.14\n",
- "r=h*math.tan(A*3.14/180.0)\n",
- "D=math.pi*r**2\n",
- "\n",
- "#Result \n",
- "print\"The required surface area is\",round(D,1),\"cm**2\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The required surface area is 1005.3 cm**2\n"
- ]
- }
- ],
- "prompt_number": 138
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 21.13 Page no 653"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "A=60 #degree\n",
- "B=1.45\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "I=1/(math.sin(A*3.14/180.0))\n",
- "Z=B/I\n",
- "\n",
- "#Result\n",
- "print\"The refractive index of the liquid is\",round(Z,3)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The refractive index of the liquid is 1.255\n"
- ]
- }
- ],
- "prompt_number": 143
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 21.14 Page no 653"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "A=1.68\n",
- "A1=1.44\n",
- "A2=90 #degree\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "K=A/A1\n",
- "C=1/K\n",
- "Q=math.asin(C)*180/3.14\n",
- "r=A2-Q\n",
- "I=A*math.sin(r*3.14/180.0)\n",
- "Q1=math.asin(I)*180/3.14\n",
- "\n",
- "#Result\n",
- "print\"Refraction index of glass fibre is\",round(Q1,0),\"degree\"\n",
- "print\"All rays having angle of incidence between 0 degree to 60 degree will suffer total internal reflection\" "
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Refraction index of glass fibre is 60.0 degree\n",
- "All rays having angle of incidence between 0 degree to 60 degree will suffer total internal reflection\n"
- ]
- }
- ],
- "prompt_number": 161
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 21.15 Page no 653"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "u1=1\n",
- "u2=1.5\n",
- "u=-10.0 #cm\n",
- "v=-40.0 #cm\n",
- "\n",
- "#Calculation\n",
- "R=(u2-u1)/((-u1/u)+(u2/v))\n",
- "\n",
- "#Result\n",
- "print\"Radius of curvature is\", R,\"cm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Radius of curvature is 8.0 cm\n"
- ]
- }
- ],
- "prompt_number": 178
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 21.16 Page no 653"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "u=-60.0 #cm\n",
- "R=25.0 #cm\n",
- "u1=1\n",
- "u2=1.5\n",
- "R1=0.25\n",
- "\n",
- "#Calculation\n",
- "v=u2/(((u2-u1)/R)+(u1/u))\n",
- "P=(u2-u1)/R1\n",
- "\n",
- "#Result\n",
- "print\"Position of the image is\", v,\"cm\"\n",
- "print\"Power of 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 refracting surface is 2.0 dioptre\n"
- ]
- }
- ],
- "prompt_number": 191
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 21.17 Page no 654 "
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "U1=1.5\n",
- "U2=1\n",
- "\n",
- "\n",
- "#Calculation\n",
- "x=(U1+U2)/(U1-U2)\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": 194
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 21.18 Page no 654"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "u1=1\n",
- "u2=1.5\n",
- "u=-10.0 #cm\n",
- "R=-5.0 \n",
- "\n",
- "#Calculation\n",
- "v=u1/((u1-u2)/R+(u2/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 -20.0 cm\n"
- ]
- }
- ],
- "prompt_number": 196
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 21.19 Page no 654"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "u=1.5\n",
- "R=-5.0 #cm\n",
- "OC=2\n",
- "\n",
- "#Calculation\n",
- "u1=R+OC\n",
- "v=1/((1-u)/R+(u/u1))\n",
- "\n",
- "#Result\n",
- "print\"Apparent position of the bubble is\", v,\"cm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Apparent position of the bubble is -2.5 cm\n"
- ]
- }
- ],
- "prompt_number": 202
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 21.20 Page no 654"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "R=-0.02 #m\n",
- "u=1.54\n",
- "v=-0.01 #m\n",
- "\n",
- "#calculation\n",
- "u1=-u/((1-u)/R-(1/v))\n",
- "\n",
- "#Result\n",
- "print\"Real depth of bubble is\", round(u1,4),\"m\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Real depth of bubble is -0.0121 m\n"
- ]
- }
- ],
- "prompt_number": 206
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 21.21 Page no 655"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "R=7.5 #cm\n",
- "u=4/3.0\n",
- "\n",
- "#Calculation\n",
- "v=1/((1-u)/R)\n",
- "\n",
- "#Result\n",
- "print\"It get focussed at\",v,\"cm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "It get focussed at -22.5 cm\n"
- ]
- }
- ],
- "prompt_number": 213
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 21.22 Page no 655"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#given\n",
- "u=1.55\n",
- "f=20 #cm\n",
- "\n",
- "#Calculation\n",
- "R=(u-1)*2*f\n",
- "\n",
- "#Result\n",
- "print\"Radius of curvature is,\",R,\"cm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Radius of curvature is, 22.0 cm\n"
- ]
- }
- ],
- "prompt_number": 215
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 21.23 Page no 655"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "u=1.5\n",
- "f=0.3\n",
- "\n",
- "#Calculation\n",
- "R=(u-1)*f\n",
- "\n",
- "#Result\n",
- "print\"Radius of curvature is\", R,\"m\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Radius of curvature is 0.15 m\n"
- ]
- }
- ],
- "prompt_number": 218
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 21.24 Page no 655"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "a=2.0\n",
- "\n",
- "#Calculation\n",
- "u=(a+1)/a\n",
- "\n",
- "#Result\n",
- "print\"Refractive index is\",u"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Refractive index is 1.5\n"
- ]
- }
- ],
- "prompt_number": 219
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 21.25 Page no 655"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "ug=1.5\n",
- "uc=1.63\n",
- "R1=20.0 #cm\n",
- "R2=-20.0\n",
- "\n",
- "#Calculation\n",
- "fair=1/((ug-1)*(1/R1-1/R2))\n",
- "ug1=ug/uc\n",
- "fc=1/((ug1-1)*(1/R1-1/R2))\n",
- "\n",
- "#Result\n",
- "print\"Focal length of the lens is\",round(fc,1),\"cm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Focal length of the lens is -125.4 cm\n"
- ]
- }
- ],
- "prompt_number": 225
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 21.26 Page no 656"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "fair=0.2 #m\n",
- "ug=1.50\n",
- "uw=1.33\n",
- "\n",
- "#Calculation\n",
- "R=1/(fair*(ug-1))\n",
- "ug1=ug/uw\n",
- "fw=1/((ug1-1)*R)\n",
- "f=fw-fair\n",
- "\n",
- "#Result\n",
- "print\"Change in focal length is\", round(f,2),\"m\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Change in focal length is 0.58 m\n"
- ]
- }
- ],
- "prompt_number": 234
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 21.27 Page no 656"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "u1=1.2\n",
- "u2=1.3\n",
- "a=13.0\n",
- "\n",
- "#Calculation\n",
- "u=u1/u2\n",
- "f=1/((-1/a)*(1/(u1-1)))\n",
- "\n",
- "#Result\n",
- "print\"Focal length is\", f,\"f\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Focal length is -2.6 f\n"
- ]
- }
- ],
- "prompt_number": 237
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 21.28 Page no 656"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "f=20.0 #cm\\\n",
- "u=12.0\n",
- "f1=-20.0\n",
- "\n",
- "#Calculation\n",
- "v=1/(1/f+1/u)\n",
- "v1=1/(1/f1+1/u)\n",
- "\n",
- "#Result\n",
- "print\"(a) Beam will converge at a point of distant\",v,\"cm\"\n",
- "print\"(b) Beam will converge at a point of distant\",v1,\"cm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a) Beam will converge at a point of distant 7.5 cm\n",
- "(b) Beam will converge at a point of distant 30.0 cm\n"
- ]
- }
- ],
- "prompt_number": 244
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 21.29 Page no 656"
- ]
- },
- {
- "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\"The position of the point is\",u,\"m\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The position of the point is 0.12 m\n"
- ]
- }
- ],
- "prompt_number": 249
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 21.30 Page no 656"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "f=10.0 #cm\n",
- "c=20 #cm\n",
- "o=5.0\n",
- "\n",
- "#Calculation\n",
- "u=-(c-o)\n",
- "u1=-(c+o)\n",
- "v1=1/((1/f)+(1/u))\n",
- "v2=1/((1/f)+(1/u1))\n",
- "v=v1-v2\n",
- "\n",
- "#Result\n",
- "print\"Length of image of needle is\", round(v,2),\"cm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Length of image of needle is 13.33 cm\n"
- ]
- }
- ],
- "prompt_number": 273
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 21.31 Page no 657"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "u=-30.0\n",
- "v=20.0\n",
- "R1=10.0\n",
- "R2=-15.0\n",
- "ug1=1.33\n",
- "\n",
- "#Calculation\n",
- "f=1/(-1/u+1/v)\n",
- "ug=(1/f+1/6.0)*6.0\n",
- "fw=1/(((ug/ug1)-1)*(1/R1-1/R2))\n",
- "\n",
- "#Result\n",
- "print\"Focal length of the lens is\", f,\"cm\"\n",
- "print\"Focal length of the lens is\",round(fw,2),\"cm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Focal length of the lens is 12.0 cm\n",
- "Focal length of the lens is 46.94 cm\n"
- ]
- }
- ],
- "prompt_number": 289
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 21.32 Page no 657"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "f=0.12\n",
- "m=-3.0\n",
- "\n",
- "#Calculation\n",
- "u=(f/m)-f\n",
- "\n",
- "#Result\n",
- "print\"Distance between object and lens is\", u,\"m\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Distance between object and lens is -0.16 m\n"
- ]
- }
- ],
- "prompt_number": 292
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 21.33 Page no 657"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "m=-19.0\n",
- "v=10\n",
- "\n",
- "#Calculation\n",
- "f=v/(-m+1)\n",
- "\n",
- "#Result\n",
- "print\"Focal length of the lens is\",f,\"m\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Focal length of the lens is 0.5 m\n"
- ]
- }
- ],
- "prompt_number": 296
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 21.34 Page no 657"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "m=4\n",
- "f=20\n",
- "\n",
- "#Calculation\n",
- "u=m+1-f\n",
- "v=f-(m*f)\n",
- "\n",
- "#Result\n",
- "print\"Object is at\", u,\"cm\"\n",
- "print\"Image is at\",v,\"cm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Object is at -15 cm\n",
- "Image is at -60 cm\n"
- ]
- }
- ],
- "prompt_number": 301
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 21.35 Page no 657"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "O=5 #cm\n",
- "u=-45.0 #cm\n",
- "v=90.0\n",
- "\n",
- "#Calculation\n",
- "f=-1/(1/u-1/v)\n",
- "I=v*O/u\n",
- "\n",
- "#Result\n",
- "print\"Focal length is\", f,\"cm\"\n",
- "print\"size of the image is\",I,\"cm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Focal length is 30.0 cm\n",
- "size of the image is -10.0 cm\n"
- ]
- }
- ],
- "prompt_number": 309
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 21.36 Page no 658"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "O=3 #cm\n",
- "u=-14.0\n",
- "f=-21.0\n",
- "\n",
- "#Calculation\n",
- "v=1/(1/f+1/u)\n",
- "I=O*v/u\n",
- "\n",
- "#Result\n",
- "print\"Image produced by the lens is\",I,\"cm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Image produced by the lens is 1.8 cm\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 21.37 Page no 658"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "R1=0.20\n",
- "R2=-0.2\n",
- "u=1.5\n",
- "u1=1.25\n",
- "\n",
- "#Calculation\n",
- "P1=(u-1)*(1/R1-1/R2)\n",
- "u2=u/u1\n",
- "P2=(u2-1)*(-1/R2-1/R2)\n",
- "P=P1/P2\n",
- "\n",
- "#Result\n",
- "print\"Ratio of power of lens is\", P"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Ratio of power of lens is 2.5\n"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 21.38 Page no 658"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "f1=15.0 #cm\n",
- "f2=30.0\n",
- "\n",
- "#Calculation\n",
- "f=1/(1/f1+1/f2)\n",
- "P=1/f\n",
- "\n",
- "#Result\n",
- "print\"Power is\",f*10**-2,\"m\"\n",
- "print\"Focal length is\",P*10**2,\"D\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Power is 0.1 m\n",
- "Focal length is 10.0 D\n"
- ]
- }
- ],
- "prompt_number": 16
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 21.39 Page no 658"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "P1=-1.5 #D\n",
- "P2=2.75\n",
- "\n",
- "#Calculation\n",
- "P=P1+P2\n",
- "f=1/P\n",
- "\n",
- "#Result\n",
- "print\"Focal length is\",f,\"m\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Focal length is 0.8 m\n"
- ]
- }
- ],
- "prompt_number": 18
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 21.40 Page no 659"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "a=0.11\n",
- "a1=0.07\n",
- "\n",
- "#Calculation\n",
- "x=a-a1\n",
- "u=(a1-0.01)/x\n",
- "\n",
- "#Result\n",
- "print\"Refractive index is\",u"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Refractive index is 1.5\n"
- ]
- }
- ],
- "prompt_number": 24
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 21.41 Page no 659"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "a=70 #Degree\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "a=1/(math.sin(a*3.14/180.0))\n",
- "a1=math.atan(a)*180/3.14\n",
- "u=1/math.sin(a1*3.14/180.0)\n",
- "\n",
- "#Result\n",
- "print\"Critical angle is\", round(u,3)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Critical angle is 1.372\n"
- ]
- }
- ],
- "prompt_number": 29
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 21.42 Page no 659"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "a=180 #degree\n",
- "b=90\n",
- "c=45\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "r=a-(b+c)\n",
- "A=1.352*math.sin(r*3.14/180.0)\n",
- "A1=math.asin(A)*180/3.14\n",
- "\n",
- "#Result\n",
- "print\"(a) Angle of incidence is sin-1[(math.sqrt(u**2-u1**2)-u1)/math.sqrt(2)]\"\n",
- "print\"(b) Angle of incidence at face AB is\",round(A1,0),\"Degree\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a) Angle of incidence is sin-1[(math.sqrt(u**2-u1**2)-u1)/math.sqrt(2)]\n",
- "(b) Angle of incidence at face AB is 73.0 Degree\n"
- ]
- }
- ],
- "prompt_number": 34
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 21.43 Page no 660"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "u=40.0 #cm\n",
- "R=10.0\n",
- "u1=1\n",
- "u2=1.5\n",
- "\n",
- "\n",
- "#Calculation\n",
- "v=u2/(((u2-u1)/R)+(u1/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 20.0 cm\n"
- ]
- }
- ],
- "prompt_number": 38
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 21.44 Page no 660"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "x=0.04\n",
- "y=8/300.0\n",
- "u=-0.04\n",
- "v=-16/500.0\n",
- "\n",
- "#Calculation\n",
- "U=x/y\n",
- "R=(1-U)/(-U/u+1/v)\n",
- "f=1/((U-1)/R)\n",
- "\n",
- "print\"(i) Refractive index is\",U\n",
- "print\"(ii) Radius of curvature is\",R,\"m\"\n",
- "print\"(iii) Focal length is\",f,\"m\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(i) Refractive index is 1.5\n",
- "(ii) Radius of curvature is -0.08 m\n",
- "(iii) Focal length is -0.16 m\n"
- ]
- }
- ],
- "prompt_number": 53
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 21.46 Page no 661"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "OI=90.0 #cm\n",
- "O1O2=20\n",
- "x=35 #cm\n",
- "\n",
- "#Calculation\n",
- "f=(OI-x)*x/((OI-x)+x)\n",
- "\n",
- "#Result\n",
- "print\"Focal length of the lens is\", round(f,1),\"cm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Focal length of the lens is 21.4 cm\n"
- ]
- }
- ],
- "prompt_number": 71
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 21.47 Page no 662"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "m=-4\n",
- "a=1.5\n",
- "u=0.3\n",
- "\n",
- "#Calculation\n",
- "x=a/(-m+1)\n",
- "v=a-u\n",
- "f=1/(1/u+1/v)\n",
- "\n",
- "#Result\n",
- "print\"Focal length of the lens is\", f,\"m\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Focal length of the lens is 0.24 m\n"
- ]
- }
- ],
- "prompt_number": 61
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 21.48 Page no 662"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "P1=8.0 #dioptre\n",
- "P2=2.0\n",
- "\n",
- "#Calculation\n",
- "f1=1/P1\n",
- "f2=1/P2\n",
- "\n",
- "#Result\n",
- "print\"Focal length of lens 1 is\",f1,\"m\"\n",
- "print\"Focal length of lens 2 is\",f2,\"m\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Focal length of lens 1 is 0.125 m\n",
- "Focal length of lens 2 is 0.5 m\n"
- ]
- }
- ],
- "prompt_number": 66
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/modern_physics_by_Satish_K._Gupta/chap22.ipynb b/modern_physics_by_Satish_K._Gupta/chap22.ipynb deleted file mode 100755 index 60d6e4e2..00000000 --- a/modern_physics_by_Satish_K._Gupta/chap22.ipynb +++ /dev/null @@ -1,531 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:168ff39597791c08930937d1b2aac6af790d1f198c96130c4b12669d6560fe5b"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 22 Dispersion"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 22.1 Page no 686"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "A=60 #Degree\n",
- "i=49\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "r=A/2.0\n",
- "u=math.sin(i*3.14/180.0)/math.sin(r*3.14/180.0)\n",
- "\n",
- "#Result\n",
- "print \"Refractive index is\",round(u,2)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Refractive index is 1.51\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 22.2 Page no 686"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "u=1.5\n",
- "A=60 #Degree\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "a1=A/2.0\n",
- "a=u*math.sin(a1*3.14/180.0)\n",
- "a2=math.asin(a)*180/3.14\n",
- "X=(a2*2)-A\n",
- "i=(A+X)/2.0\n",
- "\n",
- "#Result\n",
- "print\"Angle of deviation is\", round(X,1),\"degree\"\n",
- "print\"Angle of incidence is\",round(i,1),\"degree\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Angle of deviation is 37.2 degree\n",
- "Angle of incidence is 48.6 degree\n"
- ]
- }
- ],
- "prompt_number": 13
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 22.3 Page no 686"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "A=60 #Degree\n",
- "ug=1.53\n",
- "uw=1.33\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "ug1=ug/uw\n",
- "a2=A/2.0\n",
- "a1=ug1*math.sin(a2*3.14/180.0)\n",
- "a2=math.asin(a1)*180/3.14\n",
- "d=(a2*2)-A\n",
- "\n",
- "#Result\n",
- "print\"Angle of minimum deviation is\", round(d,1),\"Degree\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Angle of minimum deviation is 10.2 Degree\n"
- ]
- }
- ],
- "prompt_number": 20
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 22.4 Page no 686"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "ug=1.6\n",
- "uw=1.33\n",
- "i=40 #Degree\n",
- "A=60\n",
- "\n",
- "#Calculation\n",
- "ug1=ug/uw\n",
- "r1=math.asin(math.sin(i*3.14/180.0)/ug1)*180/3.14\n",
- "r2=A-r1\n",
- "e=math.asin(ug1*math.sin(r2*3.14/180.0))*180/3.14\n",
- "d=i+e-A\n",
- "\n",
- "#Result\n",
- "print\"Angle of deviation is\", round(d,0),\"Degree\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Angle of deviation is 14.0 Degree\n"
- ]
- }
- ],
- "prompt_number": 29
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 22.5 Page no 686"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "A=60 #degree\n",
- "ur=1.622\n",
- "uv=1.663\n",
- "\n",
- "#Calculation\n",
- "a=A*(uv-ur)\n",
- "\n",
- "#Result\n",
- "print\"Angle of dispersion between red and violet color is\",a,\"degree\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Angle of dispersion between red and violet color is 2.46 degree\n"
- ]
- }
- ],
- "prompt_number": 32
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 22.6 Page no 686"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "A=60 #Degree\n",
- "ab=53\n",
- "ar=51\n",
- "a=52.0\n",
- "\n",
- "#Calculation\n",
- "w=(ab-ar)/a\n",
- "\n",
- "#Result\n",
- "print\"Dispersive power is\",round(w,4)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Dispersive power is 0.0385\n"
- ]
- }
- ],
- "prompt_number": 35
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 22.7 Page no 686"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "ub=1.522\n",
- "ur=1.514\n",
- "ub1=1.662\n",
- "ur1=1.644\n",
- "\n",
- "#Calculation\n",
- "u=(ub+ur)/2.0\n",
- "w=(ub-ur)/(u-1)\n",
- "u1=(ub1+ur1)/2.0\n",
- "w1=(ub1-ur1)/(u1-1)\n",
- "\n",
- "#Result\n",
- "print\"Dispersive power of crown glass is\",round(w,4)\n",
- "print\"Dispersive power of flint galss is\",round(w1,4)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Dispersive power of crown glass is 0.0154\n",
- "Dispersive power of flint galss is 0.0276\n"
- ]
- }
- ],
- "prompt_number": 43
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 22.8 Page no 687"
- ]
- },
- {
- "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\"Refractive index is\",round(u,3)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Refractive index is 1.645\n"
- ]
- }
- ],
- "prompt_number": 46
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 22.9 Page no 687"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "uv=1.523\n",
- "ur=1.515\n",
- "A=5 #Degree\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=-(5*(u-1))/(u1-1)\n",
- "\n",
- "#Result\n",
- "print\"Angle of flint glass prism is\",round(A1,2),\"Degree\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Angle of flint glass prism is -3.88 Degree\n"
- ]
- }
- ],
- "prompt_number": 49
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 22.10 Page no 687"
- ]
- },
- {
- "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=4.2 #degree\n",
- "\n",
- "#Calculation\n",
- "A=-w1*A1*(u1-1)/(w*(u-1))\n",
- "d=-(A*(u-1)+A1*(u1-1))\n",
- "\n",
- "#Result\n",
- "print\"Angle of prism is\", d,\"Degree\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Angle of prism is 3.12 Degree\n"
- ]
- }
- ],
- "prompt_number": 56
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 22.11 Page no 687"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "A=40 #degree\n",
- "u=1.54\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "i=math.asin(u*math.sin(A*3.14/180.0))*180/3.14\n",
- "\n",
- "#Result\n",
- "print\"Angle of incidence is\",round(i,2),\"Degree\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Angle of incidence is 81.72 Degree\n"
- ]
- }
- ],
- "prompt_number": 59
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 22.12 Page no 687"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "u=1.524\n",
- "A=60 #degree\n",
- "C=41 #degree\n",
- "\n",
- "#calculation\n",
- "import math\n",
- "C1=1/u\n",
- "A1=math.asin(C1)*180/3.14\n",
- "r1=A-C\n",
- "i=u*math.sin(r1*3.14/180.0)\n",
- "A2=math.asin(i)*180/3.14\n",
- "\n",
- "#Result\n",
- "print\"Angle of incidence is\", round(A2,2),\"Degree\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Angle of incidence is 29.75 Degree\n"
- ]
- }
- ],
- "prompt_number": 84
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 22.13 Page no 688"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "u=0.015\n",
- "w=0.03\n",
- "u1=0.022\n",
- "w1=0.05\n",
- "a=2 #degree\n",
- "\n",
- "#Calculation\n",
- "A=-u1/u\n",
- "A1=a/(A/2.0+u1/w1)\n",
- "a2=A*A1\n",
- "print\"Angle of 1st prism is\", round(A1,2),\"Degree\"\n",
- "print\"Angle of 2nd prism is\",a2,\"Degree\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Angle of 1st prism is -6.82 Degree\n",
- "Angle of 2nd prism is 10.0 Degree\n"
- ]
- }
- ],
- "prompt_number": 98
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/modern_physics_by_Satish_K._Gupta/chap23.ipynb b/modern_physics_by_Satish_K._Gupta/chap23.ipynb deleted file mode 100755 index 3565eff3..00000000 --- a/modern_physics_by_Satish_K._Gupta/chap23.ipynb +++ /dev/null @@ -1,641 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:fd3a71d11d7e73e323877019a1c5f0e92d8e8b30b7f927fde2bcadd4742b58c1"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 23 Optical instruments"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 23.1 Page no 704"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "f=5.0 #cm\n",
- "D=25\n",
- "\n",
- "#Calculation\n",
- "M=1+(D/f)\n",
- "\n",
- "#Result\n",
- "print\"Magnifying power is\",M"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Magnifying power is 6.0\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 23.2 Page no 704"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "f=6.25 #cm\n",
- "v=-25\n",
- "\n",
- "#Calculation\n",
- "u=v*f/(v-f)\n",
- "M=1+(D/f)\n",
- "M1=D/f\n",
- "\n",
- "#Result\n",
- "print\"(i) The distance of the object from the lens is\", M\n",
- "print\"(ii) Anguar magnification is\",M1"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(i) The distance of the object from the lens is 5.0\n",
- "(ii) Anguar magnification is 4.0\n"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 23.3 Page no 704"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "u=-9\n",
- "f=10 #cm\n",
- "v1=-25.0\n",
- "\n",
- "#Calculation\n",
- "v=u*f/(u+f)\n",
- "M=v/u\n",
- "u=-v1*f/(v1-f)\n",
- "M1=1-(v1/f)\n",
- "\n",
- "#Result\n",
- "print\"(i) Magnification is\", M,\"\\n Size of each square in the image will appear as 100 mm**2\"\n",
- "print\"(ii) Magnification is\",M1"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(i) Magnification is 10 \n",
- " Size of each square in the image will appear as 100 mm**2\n",
- "(ii) Magnification is 3.5\n"
- ]
- }
- ],
- "prompt_number": 15
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 23.4 Page no 704"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "v=-25\n",
- "f=10.0\n",
- "\n",
- "#Calculation\n",
- "u=-v*f/(f-v)\n",
- "m=-v/u\n",
- "M=1-(v/f)\n",
- "\n",
- "#Result\n",
- "print\"(a) Distance is\", round(u,2),\"cm\"\n",
- "print\"(b) Magnification is\",m\n",
- "print\"(c) Magnification is\",M"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a) Distance is 7.14 cm\n",
- "(b) Magnification is 3.5\n",
- "(c) Magnification is 3.5\n"
- ]
- }
- ],
- "prompt_number": 28
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 23.5 Page no 705"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "M=-30\n",
- "fe=5 #cm\n",
- "D=25 #cm\n",
- "\n",
- "#Calculation\n",
- "m0=M/(1+(D/fe))\n",
- "\n",
- "#Result\n",
- "print\"Magnification produced is\",m0"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Magnification produced is -5\n"
- ]
- }
- ],
- "prompt_number": 29
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 23.6 Page no 705"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "f0=4\n",
- "u0=-6\n",
- "D=25\n",
- "fe=6.0\n",
- "\n",
- "#Calculation\n",
- "v0=f0*u0/(f0+u0)\n",
- "M=v0/u0*(1+(D/fe))\n",
- "\n",
- "#Result\n",
- "print\"Magnification produceed by the microscope is\", round(M,2)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Magnification produceed by the microscope is -10.33\n"
- ]
- }
- ],
- "prompt_number": 34
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 23.7 Page no 705"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "L=20\n",
- "fe=5\n",
- "f0=1.0 \n",
- "\n",
- "#Calculation\n",
- "v01=L-fe\n",
- "u0=v01*f0/(f0-v01)\n",
- "M=v01*D/(u0*fe)\n",
- "\n",
- "#Result\n",
- "print\"Magnification power is\", M,\"\\nThe -ve sigh shows that the image formed is inverted and real\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Magnification power is -70.0 \n",
- "The -ve sigh shows that the image formed is inverted and real\n"
- ]
- }
- ],
- "prompt_number": 41
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 23.8 Page no 705"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "f=36 #cm\n",
- "M=-8\n",
- "\n",
- "#Calculation\n",
- "fe=-f/(M-1)\n",
- "f0=-M*fe\n",
- "\n",
- "#Result\n",
- "print\"Focal length of the lenses is\", fe,\"cm and\",f0,\"cm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Focal length of the lenses is 4 cm and 32 cm\n"
- ]
- }
- ],
- "prompt_number": 46
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 23.9 Page no 705"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "f=75 #cm\n",
- "fe=5.0\n",
- "D=25 \n",
- "\n",
- "#Calculation\n",
- "M=-f/fe*(1+(fe/D))\n",
- "\n",
- "#Result\n",
- "print\"Magnying power is\",M"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Magnying power is -18.0\n"
- ]
- }
- ],
- "prompt_number": 49
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 23.10 Page no 705"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "f0=140\n",
- "fe=5\n",
- "D=25.0\n",
- "\n",
- "#Calculation\n",
- "M=-f0/fe\n",
- "M1=-(f0/fe)*(1+(fe/D))\n",
- "\n",
- "#Result\n",
- "print\"(a) Magnefying power, when telescope is in normal adjustment is\", M\n",
- "print\"(b) Magnifying power, when image is formed at least distance of distinct vision is is\",M1"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a) Magnefying power, when telescope is in normal adjustment is -28\n",
- "(b) Magnifying power, when image is formed at least distance of distinct vision is is -33.6\n"
- ]
- }
- ],
- "prompt_number": 54
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 23.11 Page no 706"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "f0=1 #m\n",
- "fe=0.05\n",
- "I=0.92*10**-2\n",
- "u0=38*10**-7\n",
- "fe1=5.0 #cm\n",
- "ve=-25\n",
- "\n",
- "#Calculation\n",
- "d=I*u0/f0\n",
- "L=f0+fe\n",
- "ue=ve*fe1/(fe1-ve)\n",
- "L1=(f0*10**2)-ue\n",
- "\n",
- "#Result\n",
- "print\"(i) Distance between two lenses is\", L*10**2,\"cm\"\n",
- "print\"(ii) Distance when final image at 25 cm from the eye is\",round(L1,2),\"cm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(i) Distance between two lenses is 105.0 cm\n",
- "(ii) Distance when final image at 25 cm from the eye is 104.17 cm\n"
- ]
- }
- ],
- "prompt_number": 71
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 23.12 Page no 706"
- ]
- },
- {
- "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 the eye piece is\",fe,\"cm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Focal length of the eye piece is 3.0 cm\n"
- ]
- }
- ],
- "prompt_number": 74
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 23.13 Page no 706"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "d=0.8 #cm\n",
- "D=508\n",
- "\n",
- "#Calculation\n",
- "B=D**2/d**2\n",
- "\n",
- "#Result\n",
- "print\"The telescope can see a star \",B,\"time sfarther than the faintest star that can be seen with naked eye.\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The telescope can see a star 403225.0 time sfarther than the faintest star that can be seen with naked eye.\n"
- ]
- }
- ],
- "prompt_number": 76
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 23.14 Page no 706"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "ve=-25 #cm\n",
- "fe=6.25\n",
- "a=15\n",
- "f0=2\n",
- "\n",
- "#Calculation\n",
- "ue=-ve*fe/(ve-fe)\n",
- "v0=a+ue\n",
- "u0=f0*v0/(f0-v0)\n",
- "v01=a-fe\n",
- "u01=v01*f0/(f0-v01)\n",
- "\n",
- "#Result\n",
- "print\"(a) Distance when final image at least distance of distinct vision is\", u0,\"cm\"\n",
- "print\"(b) Distance when final image is at infinity is\",round(u01,2),\"cm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a) Distance when final image at least distance of distinct vision is -2.5 cm\n",
- "(b) Distance when final image is at infinity is -2.59 cm\n"
- ]
- }
- ],
- "prompt_number": 87
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 23.15 Page no 707"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "a=10\n",
- "D=24\n",
- "fe=20.0\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "A=a/(1+(D/fe))/100.0\n",
- "\n",
- "#Result\n",
- "print\"Angle is\", round(A,4),\"radian\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Angle is 0.0455 radian\n"
- ]
- }
- ],
- "prompt_number": 95
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 23.16 Page no 707"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "f0=-0.5 #m\n",
- "fe=0.01\n",
- "d=3.48*10**6\n",
- "r=3.8*10**8\n",
- "a1=15\n",
- "\n",
- "#Calculation\n",
- "M=f0/fe\n",
- "a=d/r\n",
- "I=a*a1\n",
- "\n",
- "#Result\n",
- "print\"(a) Angular Magnification of the telescope is\",M\n",
- "print\"(b) Diameter of the image is\", round(I,3),\"m\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a) Angular Magnification of the telescope is -50.0\n",
- "(b) Diameter of the image is 0.137 m\n"
- ]
- }
- ],
- "prompt_number": 103
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/modern_physics_by_Satish_K._Gupta/chap25.ipynb b/modern_physics_by_Satish_K._Gupta/chap25.ipynb deleted file mode 100755 index 1f9c7095..00000000 --- a/modern_physics_by_Satish_K._Gupta/chap25.ipynb +++ /dev/null @@ -1,521 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:caf7ddcffb734a0e87ae0e90d77de524bc76c7023aae29373fc9035427092542"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 25 Interference of light "
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 25.1 Page no 728"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "w1=4\n",
- "w2=9.0\n",
- "a=25\n",
- "\n",
- "#Calculation\n",
- "I=w1/w2\n",
- "Imax=a/1.0\n",
- "\n",
- "#Result\n",
- "print\"Ratio of intensity is\",Imax"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Ratio of intensity is 25.0\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 25.2 Page no 728"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "l=4800*10**-10 #m\n",
- "b=0.6*10**-2\n",
- "b1=0.0045\n",
- "\n",
- "#Calculation\n",
- "a=l/b\n",
- "l1=a*b1*2\n",
- "\n",
- "#Result\n",
- "print\"Wavelenght of the light is\", l1*10**10,\"A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Wavelenght of the light is 7200.0 A\n"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 25.3 Page no 728"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "d=1.5*10**-3\n",
- "D=1 #m\n",
- "w=3.93 #mm\n",
- "D1=1.5\n",
- "\n",
- "#Calculation\n",
- "b=w/10.0\n",
- "l=(b*10**-3*d)/D\n",
- "a=(10*D1*l)/d\n",
- "\n",
- "#Result\n",
- "print\"Wavelength of the light is\", l,\"m\"\n",
- "print\"Width of 10 fringes is\",a*10**3,\"*10**-3 m\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Wavelength of the light is 5.895e-07 m\n",
- "Width of 10 fringes is 5.895 *10**-3 m\n"
- ]
- }
- ],
- "prompt_number": 18
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 25.4 Page no 729"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "d=0.02 #cm\n",
- "D=80\n",
- "l=6*10**-5\n",
- "n=5\n",
- "\n",
- "#Calculation\n",
- "y5=n*D*l/d\n",
- "\n",
- "#Result\n",
- "print\"Distance of fifth bright fringe is\",y5,\"cm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Distance of fifth bright fringe is 1.2 cm\n"
- ]
- }
- ],
- "prompt_number": 20
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 25.5 Page no 729"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "d=3*10**-4\n",
- "D=1.5\n",
- "y4=10**-2 #m\n",
- "\n",
- "#Calculation\n",
- "l=y4*d/(4*D)\n",
- "\n",
- "#Result\n",
- "print\"Wavelength of light is\",l,\"m\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Wavelength of light is 5e-07 m\n"
- ]
- }
- ],
- "prompt_number": 22
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 25.6 Page no 729"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "d=0.125*10**-3 #m\n",
- "l=4500*10**-10\n",
- "D=1\n",
- "\n",
- "#Calculation\n",
- "y2=2*D*l/d\n",
- "y21=2*y2\n",
- "\n",
- "#Result\n",
- "print\"Seperation between bright fringe on both sides of central maximum is\",y21*10**3,\"*10**-3 m\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Seperation between bright fringe on both sides of central maximum is 14.4 *10**-3 m\n"
- ]
- }
- ],
- "prompt_number": 27
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 25.7 Page no 729"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "d=0.012 #m\n",
- "D=100\n",
- "l=6000*10**-8 #cm\n",
- "n=2\n",
- "\n",
- "#Calculation\n",
- "y2=(n**2+1)*D*l/(n*d)\n",
- "\n",
- "#result\n",
- "print\"Distance of 3rd dark band is\",y2,\"cm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Distance of 3rd dark band is 1.25 cm\n"
- ]
- }
- ],
- "prompt_number": 29
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 25.8 Page no 729"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "d=2*10**-4 #m\n",
- "D=1.5\n",
- "y3=1.8*10**-2\n",
- "\n",
- "#Calculation\n",
- "l=2*y3*d/(7*D)\n",
- "\n",
- "#Result\n",
- "print\"Wavelength of the light is\", round(l*10**7,2)*10**-7,\"m\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Wavelength of the light is 6.86e-07 m\n"
- ]
- }
- ],
- "prompt_number": 35
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 25.9 Page no 729"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "b=2*10**-3\n",
- "l=6.0*10**-7\n",
- "u=1.33\n",
- "\n",
- "#calculation\n",
- "l1=b/l\n",
- "b2=(b*l)/(u*l)\n",
- "\n",
- "#Result\n",
- "print\"Fringe width is\",round(b2*10**3,2),\"mm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Fringe width is 1.5 mm\n"
- ]
- }
- ],
- "prompt_number": 13
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 25.10 Page no 729"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "l=6000*10**-10 #m\n",
- "d=10**-3\n",
- "D=1\n",
- "n=10\n",
- "\n",
- "#Calculation\n",
- "a10=n*l/d\n",
- "b=(D*l)/d\n",
- "\n",
- "#Result\n",
- "print\"(a) Angular position of 10th maximum is\",a10,\"radian\"\n",
- "print\"(b) Separation of the two adjacent minima is\",b*10**3,\"mm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a) Angular position of 10th maximum is 0.006 radian\n",
- "(b) Separation of the two adjacent minima is 0.6 mm\n"
- ]
- }
- ],
- "prompt_number": 18
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 25.11 Page no 730"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "l=6000*10**-10 #m\n",
- "r=0\n",
- "n=7\n",
- "\n",
- "#Calculation\n",
- "t=n*l/2.0\n",
- "\n",
- "#Result\n",
- "print\"Difference of the film is\", t,\"m\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Difference of the film is 2.1e-06 m\n"
- ]
- }
- ],
- "prompt_number": 21
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 25.12 Page no 730"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "u=1.4\n",
- "t=10**-4 #cm\n",
- "n1=4.0\n",
- "n2=5.0\n",
- "n3=6.0\n",
- "n4=7.0\n",
- "N1=9\n",
- "N2=11\n",
- "N3=13\n",
- "\n",
- "#Calculation\n",
- "a=2*u*t*10**8\n",
- "l1=a/n1\n",
- "l2=a/n2\n",
- "l3=a/n3\n",
- "l4=a/n4\n",
- "L1=2*a/N1\n",
- "L2=2*a/N2\n",
- "L3=2*a/N3\n",
- "\n",
- "#Result\n",
- "print\"(i) Wavelength When the reflection is weak is\",l1,\"A ,\",l2,\"A ,\",round(l3,0),\"A ,\",l4,\"A\"\n",
- "print\"(ii) Wavelength when the reflection is strong is\",round(L1,0),\"A ,\",round(L2,0),\"A ,\",round(L3,0),\"A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(i) Wavelength When the reflection is weak is 7000.0 A , 5600.0 A , 4667.0 A , 4000.0 A\n",
- "(ii) Wavelength when the reflection is strong is 6222.0 A , 5091.0 A , 4308.0 A\n"
- ]
- }
- ],
- "prompt_number": 43
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 25.13 Page no 730"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "n1=10\n",
- "l1=4358*10**-10 #m\n",
- "l2=5893.0*10**-10\n",
- "\n",
- "#Calculation\n",
- "n2=n1*l1/l2\n",
- "\n",
- "#Result\n",
- "print\"Number of fringes is\",round(n2,1)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Number of fringes is 7.4\n"
- ]
- }
- ],
- "prompt_number": 45
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/modern_physics_by_Satish_K._Gupta/chap26.ipynb b/modern_physics_by_Satish_K._Gupta/chap26.ipynb deleted file mode 100755 index 26f66f05..00000000 --- a/modern_physics_by_Satish_K._Gupta/chap26.ipynb +++ /dev/null @@ -1,239 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:010c48e259f2faa6331c5ac2dae668d950f9fc1e120bf60a20ed01cae899383a"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 26 Diffraction of light"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 26.1 Page no 747"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "a=0.5*10**-3 #m\n",
- "l=6000*10**-10 \n",
- "D=2\n",
- "\n",
- "#Calculation\n",
- "d=2*D*l/a\n",
- "\n",
- "#Result\n",
- "print\"Distance between the two dark bands is\",d*10**3,\"mm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Distance between the two dark bands is 4.8 mm\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 26.2 Page no 747"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "l=5900*10**-10 #m\n",
- "a=11.8*10**-7\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "a=l/a\n",
- "A=math.asin(a)*180/3.14\n",
- "A1=2*A\n",
- "\n",
- "#Result\n",
- "print\"Angular position is\",round(A1,0),\"degree\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Angular position is 60.0 degree\n"
- ]
- }
- ],
- "prompt_number": 10
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 26.3 Page no 747 "
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "l=650*10**-9\n",
- "a1=30 #degree\n",
- "\n",
- "#Calculation\n",
- "a=l/math.sin(a1*3.14/180.0)\n",
- "a2=3*l/math.sin(a1*3.14/180.0)\n",
- "\n",
- "#Result\n",
- "print\"(a) Value of a when the 1st min. falls at 30 degree is\",round(a*10**6,1)*10**-6,\"m\"\n",
- "print\"(b) Value of a when the 1st max. falls at 30 degree is\",round(a2*10**6,2)*10**-6,\"m\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a) Value of a when the 1st min. falls at 30 degree is 1.3e-06 m\n",
- "(b) Value of a when the 1st max. falls at 30 degree is 3.9e-06 m\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 26.4 Page no 747"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "l=600.0*10**-9 #m\n",
- "a=2*10**-3 \n",
- "\n",
- "#Calculation\n",
- "Z=a**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": 19
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 26.5 Page no 747"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "d=4.6*10**-6 #rad\n",
- "l=5460*10**-10 #m\n",
- "\n",
- "#Calculation\n",
- "D=1.22*l/d\n",
- "\n",
- "#Result\n",
- "print\"Aperture of the objective of the telescope is\",round(D,4),\"m\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Aperture of the objective of the telescope is 0.1448 m\n"
- ]
- }
- ],
- "prompt_number": 24
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 26.6 Page no 748"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "lr=660\n",
- "\n",
- "#Calculation\n",
- "l1=2*lr/3.0\n",
- "\n",
- "#Result\n",
- "print\"Value of lembda is\",l1,\"nm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Value of lembda is 440.0 nm\n"
- ]
- }
- ],
- "prompt_number": 26
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/modern_physics_by_Satish_K._Gupta/chap27.ipynb b/modern_physics_by_Satish_K._Gupta/chap27.ipynb deleted file mode 100755 index 9edfe788..00000000 --- a/modern_physics_by_Satish_K._Gupta/chap27.ipynb +++ /dev/null @@ -1,135 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:c77b859962fb938dcd11a74749cd47ac07047d0f8605a1cc5e5320189808ac31"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 27 Polarisation of light "
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 27.1 Page no 758"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "u=1.536\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "p=math.atan(u)*180/3.14\n",
- "r=90-p\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 33.0 degree\n"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 27.2 Page no 758"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "p=60 #degree\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "u=math.tan(p*3.14/180.0)\n",
- "C=1/u\n",
- "C1=math.asin(C)*180/3.14\n",
- "\n",
- "#Result\n",
- "print\"Critical angle is\", round(C1,1),\"Degree\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Critical angle is 35.3 Degree\n"
- ]
- }
- ],
- "prompt_number": 24
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 27.3 Page no 758"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "a=60 #degree\n",
- "b=100\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "a1=90-a\n",
- "A=a/2.0\n",
- "I=(math.cos(A*3.14/180.0)**2)/2.0\n",
- "I0=b*I\n",
- "\n",
- "#result\n",
- "print\"Unpolarised light is\", round(I0,1),\"%\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Unpolarised light is 37.5 %\n"
- ]
- }
- ],
- "prompt_number": 30
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/modern_physics_by_Satish_K._Gupta/chap28.ipynb b/modern_physics_by_Satish_K._Gupta/chap28.ipynb deleted file mode 100755 index c53c8334..00000000 --- a/modern_physics_by_Satish_K._Gupta/chap28.ipynb +++ /dev/null @@ -1,578 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:487f66e71ac79d9579f4a8e998f21bc6fadb0986ba14ee5d89b77699b48cadee"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 28 Particle nature of radiation"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 28.1 Page no 801"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "l=4950.0*10**-10\n",
- "h=6.6*10**-34 #js\n",
- "c=3*10**8 #m/s\n",
- "\n",
- "#Calculation\n",
- "E=(h*c/l)/(1.6*10**-19)\n",
- "\n",
- "#Result\n",
- "print\"Energy of each photon is\",E,\"eV\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Energy of each photon is 2.5 eV\n"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 28.2 Page no 801"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "E=6.62 #J\n",
- "v=10**12 #Hz\n",
- "h=6.62*10**-34\n",
- "\n",
- "#Calculation\n",
- "E1=h*v\n",
- "n=E/(h*v)\n",
- "\n",
- "#Result\n",
- "print\"Number of photons is\",n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Number of photons is 1e+22\n"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 28.3 Page no 801"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "h=6.62*10**-34 #js\n",
- "c=3*10**8 #m/s\n",
- "Iev=1.6*10**-19 #J\n",
- "V=10**6 #hz\n",
- "V1=5890*10**-10 #m\n",
- "L=10.0**-10 #m\n",
- "\n",
- "#Calculation\n",
- "E=h*v\n",
- "E1=E/Iev\n",
- "A=(h*c)/V1\n",
- "A1=A/Iev\n",
- "B=(h*c)/L\n",
- "B1=B/Iev\n",
- "\n",
- "#Result\n",
- "print\"(i) The energy of a photon with frequency 1000 Khz is\",round(E1*10**3,2),\"10**-9\",\"ev\" \n",
- "print\"(ii) The energy of a photon when wavelength is 5890A is\",round(A1,2) ,\"eV\"\n",
- "print\"(iii) The energy of a photon when wavelength is 1 A is\", B1,\"eV\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(i) The energy of a photon with frequency 1000 Khz is 4.14 10**-9 ev\n",
- "(ii) The energy of a photon when wavelength is 5890A is 2.11 eV\n",
- "(iii) The energy of a photon when wavelength is 1 A is 12412.5 eV\n"
- ]
- }
- ],
- "prompt_number": 38
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 28.4 Page no 801"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "l=0.66*10**-10 #m\n",
- "h=6.62*10**-34 #Js\n",
- "c=3*10**8 #m/s\n",
- "\n",
- "#Calculation\n",
- "E=((h*c)/l)/(1.6*10**-19)\n",
- "\n",
- "#Result\n",
- "print\"(a) Maximum energy of photon is\",round(E*10**-3,2),\"Kev\"\n",
- "print\"(b) To produce electrons of energy 18.81 Kev,accelerating potential of 18.81 KV\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a) Maximum energy of photon is 18.81 Kev\n",
- "(b) To produce electrons of energy 18.81 Kev,accelerating potential of 18.81 KV\n"
- ]
- }
- ],
- "prompt_number": 45
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 28.5 Page no 801"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "E=10.2*10**9 #ev\n",
- "h=6.62*10**-34\n",
- "c=3*10**8\n",
- "\n",
- "#Calculation\n",
- "E1=(E/2.0)*(1.6*10**-19)\n",
- "l=(h*c)/E1\n",
- "\n",
- "#Result\n",
- "print\"Wavelength is\",round(l*10**16,3),\"10**-16\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Wavelength is 2.434 10**-16\n"
- ]
- }
- ],
- "prompt_number": 60
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 28.6 Page no 802"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "L=3500.0*10**-10 #m\n",
- "h=6.62*10**-34\n",
- "c=3*10**8\n",
- "\n",
- "#Calculation\n",
- "Hv=h*c/(L*(1.6*10**-19))\n",
- "\n",
- "#Result\n",
- "print\"Energy of incident light is\", round(Hv,3),\"eV \\nMetal B will yield photoelectrons.\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Energy of incident light is 3.546 eV \n",
- "Metal B will yield photoelectrons.\n"
- ]
- }
- ],
- "prompt_number": 75
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 28.7 Page no 802"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "l=6000.0*10**-10 #m\n",
- "h=6.62*10**-34 #Js\n",
- "e=1.6*10**-19\n",
- "c=3*10**8\n",
- "\n",
- "#Calculation\n",
- "v0=c/l\n",
- "w=(h*v0)/e\n",
- "\n",
- "#Result\n",
- "print\"(i) Threshold frequency is\", v0,\"Hz\"\n",
- "print\"(ii) Work function is\",round(w,3),\"eV\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(i) Threshold frequency is 5e+14 Hz\n",
- "(ii) Work function is 2.069 eV\n"
- ]
- }
- ],
- "prompt_number": 83
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 28.8 Page no 802"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "w=1.2*1.6*10**-19\n",
- "l=5000.0*10**-10\n",
- "c=3*10**8\n",
- "h=6.62*10**-34\n",
- "e=1.6*10**-19\n",
- "\n",
- "#Calculation\n",
- "A=((h*c)/l)-w\n",
- "V0=A/e\n",
- "\n",
- "#Result\n",
- "print\"Stopping potential is\", round(V0,2),\"V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Stopping potential is 1.28 V\n"
- ]
- }
- ],
- "prompt_number": 90
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 28.9 Page no 802"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "h=6.6*10**-34 #Js\n",
- "e=1.6*10**-19\n",
- "l=500.0*10**-9 #m\n",
- "\n",
- "#Calculation\n",
- "w=2*e\n",
- "l0=(h*c)/w\n",
- "E=((h*c)/l)-w\n",
- "v0=E/e\n",
- "\n",
- "#Result\n",
- "print\"(i) The threshold wavelength is\", l0*10**9,\"nm\"\n",
- "print\"(ii) Maximum energy is\",E*10**19,\"*10**-19 J\"\n",
- "print\"(iii) The stopping potential is\",v0,\"V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(i) The threshold wavelength is 618.75 nm\n",
- "(ii) Maximum energy is 0.76 *10**-19 J\n",
- "(iii) The stopping potential is 0.475 V\n"
- ]
- }
- ],
- "prompt_number": 104
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 28.10 Page no 802"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "V0=1.3\n",
- "l=2271*10**-10\n",
- "h=6.62*10**-34\n",
- "e=1.6*10**-19\n",
- "c=3*10**8\n",
- "\n",
- "#Calculation\n",
- "w=(h*c/l)-(V0*e)\n",
- "l0=(h*c)/w\n",
- "\n",
- "#Result\n",
- "print\"Work function of metal is\", round(l0*10**10,0),\"A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Work function of metal is 2980.0 A\n"
- ]
- }
- ],
- "prompt_number": 109
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 28.11 Page no 803"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "h=6.62*10**-34\n",
- "c=3*10**8\n",
- "l=640.2*10**-9\n",
- "e=1.6*10**-19\n",
- "V0=0.54\n",
- "l1=427.2*10**-9\n",
- "\n",
- "#Calculation\n",
- "w=(h*c/l)-(e*V0)\n",
- "v0=(((h*c)/l1)-w)/e\n",
- "\n",
- "#Result\n",
- "print\"New stopping voltage is\", round(v0,3),\"V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "New stopping voltage is 1.507 V\n"
- ]
- }
- ],
- "prompt_number": 115
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 28.12 Page no 803"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "l=5600.0*10**-10 #m\n",
- "h=6.625*10**-34\n",
- "c=3*10**8\n",
- "a=5\n",
- "\n",
- "#Calculation\n",
- "E=h*c/l\n",
- "n=a/E\n",
- "\n",
- "#Result\n",
- "print\"Number of visible light photon is\", round(n*10**-19,2)*10**-19"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Number of visible light photon is 1.41e-19\n"
- ]
- }
- ],
- "prompt_number": 121
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 28.13 Page no 803"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "E=10**4 #W\n",
- "l=500.0 #m\n",
- "h=6.62*10**-34\n",
- "A=0.4*10**-14 #m**2\n",
- "i=10**-10 #W/m**2\n",
- "v=6*10**14\n",
- "\n",
- "#Calculation\n",
- "E1=E*l/(h*c)\n",
- "E2=h*v\n",
- "n=A/E2\n",
- "\n",
- "#Result\n",
- "print\"(i) Number of photons emitted is\",round(E1*10**-31,3)*10**31\n",
- "print\"(ii) Number of photons is\", round(n*10**-4,0),\"*10**4\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(i) Number of photons emitted is 2.518e+31\n",
- "(ii) Number of photons is 1.0 *10**4\n"
- ]
- }
- ],
- "prompt_number": 132
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 28.14 Page no 803"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "A=2*10**-4 #m**2\n",
- "An=10.0**-20\n",
- "n1=5\n",
- "i=10**-5 #W/m**2\n",
- "\n",
- "#Calculation\n",
- "n=(A*n1)/An\n",
- "E=i*A\n",
- "E1=(E/n)/e\n",
- "t=2/E1\n",
- "\n",
- "#Result\n",
- "print\"Time required for the photoelectric emission is\",t*10**-7,\"*10**7 S\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Time required for the photoelectric emission is 1.6 *10**7 S\n"
- ]
- }
- ],
- "prompt_number": 145
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/modern_physics_by_Satish_K._Gupta/chap3.ipynb b/modern_physics_by_Satish_K._Gupta/chap3.ipynb deleted file mode 100755 index 7c6ed32c..00000000 --- a/modern_physics_by_Satish_K._Gupta/chap3.ipynb +++ /dev/null @@ -1,783 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:c3ec75ccee40d0eb91752b49524186dbe794af4f0e24a564bb72af8d09f81af1"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 3 Electric potential"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 3.1 Page no 65"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "q=4.0 #c\n",
- "Va=-10 #V\n",
- "Wab=100 #joule\n",
- "\n",
- "#Calculation\n",
- "V=(Wab/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 15.0 Volts\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 3.2 Page no 65"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "Z=79\n",
- "e=1.6*10**-19 #C\n",
- "r=6.6*10**-15 #m\n",
- "m=9*10**9\n",
- "\n",
- "#Calculation\n",
- "q=Z*e\n",
- "V=(m*q)/r\n",
- "\n",
- "#Result\n",
- "print\"Electric potential at the surface of a gold nucleus is\", round(V*10**-7,3),\"*10**7 V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Electric potential at the surface of a gold nucleus is 1.724 *10**7 V\n"
- ]
- }
- ],
- "prompt_number": 10
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 3.3 Page no 65"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "q=4*10**-7 #C\n",
- "r=0.09 #m\n",
- "m=9*10**9\n",
- "a=2*10**-9\n",
- "\n",
- "#Calculation\n",
- "Vp=(m*q)/r\n",
- "W=a*Vp\n",
- "\n",
- "#Result\n",
- "print\"(a) Potential at point P is\", Vp*10**-4,\"*10**4 V\"\n",
- "print\"(b) Work done is\", W,\"J\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a) Potential at point P is 4.0 *10**4 V\n",
- "(b) Work done is 8e-05 J\n"
- ]
- }
- ],
- "prompt_number": 17
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 3.4 Page no 66"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "q=20*10**-6 #C\n",
- "r1=0.1 #m\n",
- "r2=0.05 #m\n",
- "m=9*10**9\n",
- "e=1.6*10**-19\n",
- "\n",
- "#Calculation\n",
- "Va=(m*q)/r1\n",
- "Vb=(m*q)/r2\n",
- "Wab=-(Va-Vb)*e\n",
- "\n",
- "#Result\n",
- "print\"Potential at point A is\", Va*10**-6,\"*10**6 V\"\n",
- "print\"Potential at point b is\", Vb*10**-6,\"*10**6 V\"\n",
- "print\"Work done is\",Wab,\"J\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Potential at point A is 1.8 *10**6 V\n",
- "Potential at point b is 3.6 *10**6 V\n",
- "Work done is 2.88e-13 J\n"
- ]
- }
- ],
- "prompt_number": 29
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 3.5 Page no 66"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "m=9.0*10**9\n",
- "r=0.5\n",
- "a=16\n",
- "\n",
- "#Calculation\n",
- "q=(a*r)/m\n",
- "\n",
- "#Result\n",
- "print\"Magnitude of charge is\", round(q*10**10,2)*10**-10,\"C\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Magnitude of charge is 8.89e-10 C\n"
- ]
- }
- ],
- "prompt_number": 35
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 3.6 Page no 66"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "q1=2*10**-6 #c\n",
- "q2=-2*10**-6\n",
- "q3=-3*10**-6\n",
- "q4=6*10**-6\n",
- "r=1.0 #m\n",
- "m=9*10**9\n",
- "\n",
- "#Calculation\n",
- "V=m*(q1+q2+q3+q4)/r\n",
- "\n",
- "#Result\n",
- "print\"Potential at the centre of the square is\",V*10**-4,\"*10**4 V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Potential at the centre of the square is 2.7 *10**4 V\n"
- ]
- }
- ],
- "prompt_number": 41
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 3.7 Page no 66"
- ]
- },
- {
- "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",
- "V=(6*m*q)/r\n",
- "\n",
- "#Result\n",
- "print\"Potential at the centre of the hexagon is\", V*10**-6,\"*10**6 V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Potential at the centre of the hexagon is 2.7 *10**6 V\n"
- ]
- }
- ],
- "prompt_number": 45
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 3.9 Page no 67"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "r=0.1 #m\n",
- "q=200*10**-6 #C\n",
- "m=9*10**9\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 1.8 *10**7 V\n"
- ]
- }
- ],
- "prompt_number": 50
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 3.11 Page no 67"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "a=0.2 #m\n",
- "m=9*10**9\n",
- "q1=2*10**-9\n",
- "q2=4*10**-9\n",
- "q3=8*10**-9\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "V=m*((q1/a)+(q2/(a*math.sqrt(2)))+(q3/a))\n",
- "b=0.1*math.sqrt(2)\n",
- "V0=m*((q1/b)+(q2/b)+(q3/b))\n",
- "W=q*(V0-V)\n",
- "\n",
- "#Result\n",
- "print\"Required work is\",round(W*10**2,2)*10**-7,\"J\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Required work is 6.27e-07 J\n"
- ]
- }
- ],
- "prompt_number": 62
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 3.12 Page no 68"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "V=60\n",
- "r=0.05\n",
- "\n",
- "#Calculation\n",
- "E=-V/r\n",
- "\n",
- "#Result\n",
- "print\"Electric field intensity between the two plate is\", E,\"V/m\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Electric field intensity between the two plate is -1200.0 V/m\n"
- ]
- }
- ],
- "prompt_number": 66
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 3.13 Page no 68"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "e=1.6*10**-19\n",
- "m=3*10**-16 #Kg\n",
- "r=5*10**-3 #m\n",
- "g=9.8\n",
- "\n",
- "#Calculation\n",
- "q=10*e\n",
- "V=(m*g*r)/q\n",
- "\n",
- "#Result\n",
- "print round(V,2),\"V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "9.19 V\n"
- ]
- }
- ],
- "prompt_number": 71
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 3.14 Page no 68"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "q1=7*10**-6 #c\n",
- "q2=-2*10**-6 #C\n",
- "r12=0.18 #m\n",
- "m=9*10**9\n",
- "\n",
- "#Calculation\n",
- "U=m*((q1*q2)/r12)\n",
- "W=0-U\n",
- "\n",
- "#Result\n",
- "print\"(a) Electrostatic potential energy of a system is\",U,\"J\"\n",
- "print\"(b) work required is\", W,\"J\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a) Electrostatic potential energy of a system is -0.7 J\n",
- "(b) work required is 0.7 J\n"
- ]
- }
- ],
- "prompt_number": 78
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 3.15 Page no 68"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "m=9*10**9\n",
- "e=1.6*10**-19 #C\n",
- "r=5.3*10**-11 #m\n",
- "\n",
- "#Calculation\n",
- "V=(m*e)/r\n",
- "U=((m*e*-e)/r)/e\n",
- "q=2*e\n",
- "V1=(m*q)/r\n",
- "\n",
- "#Result\n",
- "print\"(a) Electric potential is\", round(V,2),\"V\"\n",
- "print\"(b) Electric potential energy is\",round(V1,2),\"V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a) Electric potential is 27.17 V\n",
- "(b) Electric potential energy is 54.34 V\n"
- ]
- }
- ],
- "prompt_number": 91
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 3.16 Page no 69"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "r12=1.5*10**-10 #m\n",
- "r13=10.0**-10\n",
- "q1=1.6*10**-19 #C\n",
- "q3=-1.6*10**-19\n",
- "m=9*10**9\n",
- "\n",
- "#Calculation\n",
- "U=m*(((q1**2)/r12)+((q1*q3)/r13)+((q1*q3)/r13))/q1\n",
- "\n",
- "#Result\n",
- "print\"Potential energy is\",U,\"ev\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Potential energy is -19.2 ev\n"
- ]
- }
- ],
- "prompt_number": 109
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 3.17 Page no 69"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "E=10**5 #N/C\n",
- "q=10**-6 #C\n",
- "a=0.02 #m\n",
- "A=1\n",
- "\n",
- "#Calculation'\n",
- "import math\n",
- "p=q*a\n",
- "t=p*E*A\n",
- "W=p*E*(math.cos(0*3.14/180.0)-math.cos(180*3.14/180.0))\n",
- "\n",
- "#Result\n",
- "print\"(i) Torque acting on dipole is\",t,\"N m\"\n",
- "print\"(ii) Work done is\", round(W,3),\"J\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(i) Torque acting on dipole is 0.002 N m\n",
- "(ii) Work done is 0.004 J\n"
- ]
- }
- ],
- "prompt_number": 126
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 3.18 Page no 69"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "m=9*10**9\n",
- "q=8*10**-3\n",
- "q0=-2*10**-9\n",
- "ra=0.03\n",
- "rb=0.04\n",
- "\n",
- "#Calculation\n",
- "W=(m*q*q0*(1/rb-1/ra))\n",
- "\n",
- "#Result\n",
- "print\"Work done is\", W,\"J\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Work done is 1.2 J\n"
- ]
- }
- ],
- "prompt_number": 129
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 3.19 Page no 69"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "qa=1.5*10**-6 #C\n",
- "qb=2.5*10**-6 \n",
- "a=0.15 #m\n",
- "m=9*10**9\n",
- "a1=0.18\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "V=((qa+qb)/a)*m\n",
- "Ea=m*qa/a**2\n",
- "Eb=m*qb/a**2\n",
- "E=Eb-Ea\n",
- "V1=((qa+qb)/a1)*m\n",
- "Ea1=m*qa/a1**2\n",
- "Ea2=m*qb/a1**2\n",
- "A=math.atan(15/10.0)*180/3.14\n",
- "A1=2*A\n",
- "A2=math.cos(A1*3.14/180.0)\n",
- "E3=math.sqrt(Ea1**2+Ea2**2+(2*Ea1*Ea2*A2))\n",
- "tb=Ea1*math.sin(A1*3.14/180.0)/(Ea2+Ea1*math.cos(A1*3.14/180.0))\n",
- "B=math.atan(tb)*180/3.14\n",
- "y=90-A\n",
- "Y=y+B\n",
- "\n",
- "#Result\n",
- "print \"(a) Net electric field is\",E*10**-5,\"*10**5 N/C\"\n",
- "print\"(b) Angle is\" ,round(Y,1),\"degree\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a) Net electric field is 4.0 *10**5 N/C\n",
- "(b) Angle is 69.4 degree\n"
- ]
- }
- ],
- "prompt_number": 160
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 3.23 Page no 71"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "q1=-1.6*10**-19 #C\n",
- "q2=1.6*10**19\n",
- "r12=0.53*10**-10 #m\n",
- "m=9*10**9\n",
- "r11=1.06*10**-10\n",
- "\n",
- "#Calculation\n",
- "U=((m*q1*q2)/r12)/q2\n",
- "K=U/2.0\n",
- "E=-(U-K)\n",
- "U1=((m*q1*q2)/r11)/q2\n",
- "E1=U-U1\n",
- "\n",
- "#Result\n",
- "print \"(a) Potential energy is\",round(U,2),\"ev\"\n",
- "print\"(b) Minimum work required is\",round(E,3) ,\"ev\"\n",
- "print\"(c) Potential energy is\",round(E1,3),\"ev\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a) Potential energy is -27.17 ev\n",
- "(b) Minimum work required is 13.585 ev\n",
- "(c) Potential energy is -13.585 ev\n"
- ]
- }
- ],
- "prompt_number": 181
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 3.24 Page no 72"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "E=10**6 #V/m\n",
- "p=10**-29 #cm\n",
- "a=6.023*10**23\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "Np=a*p\n",
- "U1=Np*E\n",
- "U2=Np*E*math.cos(60*3.14/180.0)\n",
- "U=U1-U2\n",
- "\n",
- "#Result\n",
- "print\"Heat released by the substance is\", round(U,2),\"J\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Heat released by the substance is 3.01 J\n"
- ]
- }
- ],
- "prompt_number": 188
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/modern_physics_by_Satish_K._Gupta/chap30.ipynb b/modern_physics_by_Satish_K._Gupta/chap30.ipynb deleted file mode 100755 index cf038d52..00000000 --- a/modern_physics_by_Satish_K._Gupta/chap30.ipynb +++ /dev/null @@ -1,235 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:75345a10ab4b01b2d63b7c49ca20de9a23e8e81cbab11c7f6d702383f9b2feef"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 30 Structure Of Nucleus"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 30.1 Page no 840"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "A1=197\n",
- "A2=107.0\n",
- "\n",
- "#Calculation\n",
- "R=(A1/A2)**(0.3)\n",
- "\n",
- "#Result\n",
- "print\"The ratio of the nuclear radii of the gold and silver isotope is\",round(R,3)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The ratio of the nuclear radii of the gold and silver isotope is 1.201\n"
- ]
- }
- ],
- "prompt_number": 19
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 30.2 Page no 840"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "R=1.2*10**-15 #m\n",
- "M=1.67*10**-27 #kg\n",
- "P1=10**3 #kg m**-3\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "V=1.3*math.pi*(R**3)\n",
- "P=M/V\n",
- "A=P/P1\n",
- "\n",
- "#Result\n",
- "print\"Nuclear matter is denser than water is\",round(A*10**-14,3),\"*10**14\",\"times\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Nuclear matter is denser than water is 2.366 10**14 times\n"
- ]
- }
- ],
- "prompt_number": 42
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 30.3 Page no 840"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "Mh=4.00150 #a.m.u.\n",
- "Mp=1.00728 #a.m.u.\n",
- "Mn=1.00867 #a.m.u.\n",
- "W0=931.5 #MeV\n",
- "\n",
- "#Calculation\n",
- "A=((2*Mp)+(2*Mn))-Mh\n",
- "A1=A*W0\n",
- "\n",
- "#Result\n",
- "print\"Binding energy of a-particle is\",round(A1,2),\"MeV\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Binding energy of a-particle is 28.32 MeV\n"
- ]
- }
- ],
- "prompt_number": 49
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 30.4 Page no 840"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "Mp=1.007275 #a.m.u.\n",
- "Mn=1.008665 #a.m.u.\n",
- "Mh=2.013553 #a.m.u.\n",
- "S=2.0\n",
- "U=931.5\n",
- "\n",
- "#Calculation\n",
- "A=(Mp+Mn)-Mh\n",
- "P=A/S\n",
- "W=A*U\n",
- "L=W/S\n",
- "\n",
- "#Result\n",
- "print\"The mass defect is\",A,\"a.m.u.\" \n",
- "print\"The packing fraction is\",P,\"a.m.u.\"\n",
- "print\"The binding energy of deutron is\",round(W,2),\"MeV\"\n",
- "print\"The binding energy of per nucleon of deutron is\",round(L,2),\"MeV\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The mass defect is 0.002387 a.m.u.\n",
- "The packing fraction is 0.0011935 a.m.u.\n",
- "The binding energy of deutron is 2.22 MeV\n",
- "The binding energy of per nucleon of deutron is 1.11 MeV\n"
- ]
- }
- ],
- "prompt_number": 62
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 30.5 Page no 840"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "Mh=1.007825 #a.m.u.\n",
- "Mn=1.008665 #a.m.u.\n",
- "Mp=55.934939 #a.m.u.\n",
- "Mb=208.980388 #a.m.u.\n",
- "A=56.0\n",
- "Z=26\n",
- "S=931.5\n",
- "A1=209.0\n",
- "Z1=83\n",
- "\n",
- "#Calculation\n",
- "W=A-Z\n",
- "Q=((Z*Mh+W*Mn)-Mp)*S\n",
- "R=Q/A\n",
- "W1=A1-Z1\n",
- "Q1=((Z1*Mh+W1*Mn)-Mb)*S\n",
- "R1=Q1/A1\n",
- "\n",
- "#Result\n",
- "print\"The binding energy of the nuclri of Fe is\",round(Q,2),\"MeV\"\n",
- "print\"The binding energy of the nuclei of Bi is\",round(Q1,2),\"MeV\"\n",
- "print\"Binding energy per nucleon of Fe is\",round(R,2),\"MeV\"\n",
- "print\"Binding energy per nucleon of Bi is\",round(R1,3),\"MeV\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The binding energy of the nuclri of Fe is 492.26 MeV\n",
- "The binding energy of the nuclei of Bi is 1640.26 MeV\n",
- "Binding energy per nucleon of Fe is 8.79 MeV\n",
- "Binding energy per nucleon of Bi is 7.848 MeV\n"
- ]
- }
- ],
- "prompt_number": 80
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/modern_physics_by_Satish_K._Gupta/chap31.ipynb b/modern_physics_by_Satish_K._Gupta/chap31.ipynb deleted file mode 100755 index 6135b7c0..00000000 --- a/modern_physics_by_Satish_K._Gupta/chap31.ipynb +++ /dev/null @@ -1,388 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:64e0a844a3ba33dead720cfb202f40f7dab27e1a9fab7ab4975bb30624a36195"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 31 Radioactivity"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 31.1 Page no 855"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "T=138.0 #days\n",
- "E=0.693\n",
- "N=12.5\n",
- "N0=100\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "L=E/T\n",
- "N1=N/N0\n",
- "t=(2.303*math.log10(8))/L\n",
- "\n",
- "#Result\n",
- "print\"Time is\",round(t,2),\"days\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Time is 414.16 days\n"
- ]
- }
- ],
- "prompt_number": 71
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 31.2 Page no 855"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "T=1.42*10**17 #s\n",
- "A=6.02*10**23 #mol**-1\n",
- "E=0.693\n",
- "L=238.0\n",
- "\n",
- "#Calculation\n",
- "N=A/L\n",
- "L1=E/T\n",
- "Z=L1*N\n",
- "\n",
- "#Result\n",
- "print\"Disintegrations per second occur in 1g of U**238 is\",round(Z*10**-4,3),\"*10**4 s**-1\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Disintegrations per second occur in 1g of U**238 is 1.234 *10**4 s**-1\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 31.3 Page no 855"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "K=-40 #Cl\n",
- "M=0.075 #kg mole**-1\n",
- "m=1.2*10**-6 #kg\n",
- "A=6.0*10**23\n",
- "D=170 #s**-1\n",
- "E=0.693\n",
- "\n",
- "#Calculation\n",
- "N=(A/M)*m\n",
- "L=D/N\n",
- "T=(E/L)\n",
- "O=T/31536000.0\n",
- "\n",
- "#Result\n",
- "print\"Half life of K40 atom is\",round(O*10**-9,3),\"*10**9 years\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Half life of K40 atom is 1.241 *10**9 years\n"
- ]
- }
- ],
- "prompt_number": 81
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 31.4 Page no 856"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "T=6.0 #hours\n",
- "E=0.693\n",
- "A=6.025*10**23\n",
- "W=99.0\n",
- "S=10**-12\n",
- "t=1 #hours\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "L=E/T\n",
- "N0=(A/W)*S\n",
- "R0=L*N0\n",
- "N=N0*math.exp(-L)\n",
- "R=L*N\n",
- "R1=L*N\n",
- "\n",
- "#Result\n",
- "print\"Activity at beginning is\", round(R0*10**-8,2),\"*10**8 /h\"\n",
- "print\"Activity at the end is\",round(R1*10**-8,3),\"*10**8 /h\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Activity at beginning is 7.03 *10**8 /h\n",
- "Activity at the end is 6.262 *10**8 /h\n"
- ]
- }
- ],
- "prompt_number": 15
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 31.5 Page no 856"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "t=30.0 #years\n",
- "a=16\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "l=(2.3026*math.log10(a))/t\n",
- "T=0.693/l\n",
- "\n",
- "#Result\n",
- "print\"Half life period is\", round(T,1),\"years\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Half life period is 7.5 years\n"
- ]
- }
- ],
- "prompt_number": 22
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 31.6 Page no 856"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "m1=238.05081\n",
- "m2=234.04363\n",
- "m3=4.00260\n",
- "A=931.5 #Mev\n",
- "\n",
- "#Calculation\n",
- "Ea=(m1-m2-m3)*A\n",
- "\n",
- "#Result\n",
- "print\"Kinetic energy is\",round(Ea,2),\"Mev\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Kinetic energy is 4.27 Mev\n"
- ]
- }
- ],
- "prompt_number": 25
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 31.7 Page no 856"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "m1=22.994466\n",
- "m2=22.989770\n",
- "A=931.5\n",
- "\n",
- "#Calculation\n",
- "Eb=(m1-m2)*A\n",
- "\n",
- "#Result\n",
- "print\"Maximum kinetic energy is\",round(Eb,3),\"Mev\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Maximum kinetic energy is 4.374 Mev\n"
- ]
- }
- ],
- "prompt_number": 28
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 31.8 Page no 856"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "E1=0\n",
- "E2=0.412*1.6*10**-13\n",
- "E3=1.088*1.6*10**-13\n",
- "h=6.62*10**-34\n",
- "m1=197.968233\n",
- "m2=197.966760\n",
- "a=931.5\n",
- "A=1.088\n",
- "A1=0.412\n",
- "\n",
- "#Calculation\n",
- "v1=(E3-E1)/h\n",
- "v2=(E2-E1)/h\n",
- "v3=(E3-E2)/h\n",
- "Eb1=((m1-m2)*a)-A\n",
- "Eb2=((m1-m2)*a)-A1\n",
- "\n",
- "#Result\n",
- "print\"Radiation frequencies are\",round(v1*10**-20,2)*10**20,\",\",round(v2*10**-20,3)*10**20,\"and\",round(v3*10**-20,2)*10**20\n",
- "print\"Maximum kinetic energies are\",round(Eb1,3),\"Mev and \",round(Eb2,3),\"Mev\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Radiation frequencies are 2.63e+20 , 9.96e+19 and 1.63e+20\n",
- "Maximum kinetic energies are 0.284 Mev and 0.96 Mev\n"
- ]
- }
- ],
- "prompt_number": 42
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 31.9 Page no 857"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "a=15\n",
- "b=9.0\n",
- "T=5730\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "A=a/b\n",
- "l=2.303*math.log10(A)\n",
- "t=(l*T)/0.693\n",
- "\n",
- "#Result\n",
- "print\"Approximation age of the indus valley civilization is\",round(t,2),\"years\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Approximation age of the indus valley civilization is 4224.47 years\n"
- ]
- }
- ],
- "prompt_number": 46
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/modern_physics_by_Satish_K._Gupta/chap32.ipynb b/modern_physics_by_Satish_K._Gupta/chap32.ipynb deleted file mode 100755 index 69ecaa80..00000000 --- a/modern_physics_by_Satish_K._Gupta/chap32.ipynb +++ /dev/null @@ -1,381 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:1a205c66d8c05960f2478abf5942d3dc43db8b63d7327dc1bc8decd646c4e601"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 32 Nuclear reactions"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 32.3 Page no 874"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "r1=6.015126\n",
- "r2=1.008665\n",
- "p1=4.002603\n",
- "p2=3.016049\n",
- "a=931\n",
- "\n",
- "#Calculation\n",
- "R=r1+r2\n",
- "P=p1+p2\n",
- "M=R-P\n",
- "E=M*a\n",
- "\n",
- "#Result\n",
- "print\"Energy released is\",round(E,2),\"Mev\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Energy released is 4.78 Mev\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 32.4 Page no 874"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "P=400*10**6 #J/s\n",
- "c=3.0*10**8\n",
- "\n",
- "#Calculation\n",
- "E=P*24*60*60\n",
- "m=E/c**2\n",
- "\n",
- "#Result\n",
- "print\"U235 is\",m*10**3,\"g\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "U235 is 0.384 g\n"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 32.5 Page no 874"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "A=6.023*10**23\n",
- "w=235.0\n",
- "E=200\n",
- "e=1.6*10**-13\n",
- "\n",
- "#Calculation\n",
- "E1=((A*E)/w)*e\n",
- "E2=E1/(1000.0*3600.0)\n",
- "\n",
- "#Result\n",
- "print\"Energy released is\", round(E2*10**-4,3),\"*10**4 KWh\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Energy released is 2.278 *10**4 KWh\n"
- ]
- }
- ],
- "prompt_number": 17
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 32.6 Page no 874"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "E=7.6*10**13 #J\n",
- "E1=200.0*1.6*10**-13\n",
- "h=6.023*10**23\n",
- "a=235\n",
- "\n",
- "#Calculation\n",
- "n=E/E1\n",
- "M=(a*n)/h\n",
- "\n",
- "#Result\n",
- "print\"(i) Number of uranium atom is\",n\n",
- "print\"(ii) Mass of uranium is\",round(M,2),\"g\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(i) Number of uranium atom is 2.375e+24\n",
- "(ii) Mass of uranium is 926.66 g\n"
- ]
- }
- ],
- "prompt_number": 22
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 32.7 Page no 874"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "m1=235.043933\n",
- "m2=140.917700\n",
- "m3=91.895400\n",
- "n1=1.0086651\n",
- "a=931.5\n",
- "\n",
- "#Calculation\n",
- "Q=(m1-m2-m3-2*(n1))*a\n",
- "\n",
- "#Result\n",
- "print\"Energy released is\",round(Q,2),\"Mev\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Energy released is 198.88 Mev\n"
- ]
- }
- ],
- "prompt_number": 26
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 32.8 Page no 874"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "M1=1.007825\n",
- "M2=4.002603\n",
- "M3=0.000549\n",
- "A=931.5\n",
- "S=4\n",
- "W=2\n",
- "\n",
- "#Calculation\n",
- "Q=((S*M1-M2-W*M3)*A)\n",
- "\n",
- "#Result\n",
- "print\"The energy is\",round(Q,2),\"MeV\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The energy is 25.71 MeV\n"
- ]
- }
- ],
- "prompt_number": 30
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 32.9 Page no 875"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "M1=1.008 #a.m.u.\n",
- "M2=4.004 #a.m.u.\n",
- "M3=7.016 #a.m.u.\n",
- "A=2\n",
- "W=931\n",
- "\n",
- "#Calculation\n",
- "Q=(M3+M1-A*M2)*W\n",
- "\n",
- "#Result\n",
- "print\"The initial energy of each alpha-particle is\",Q,\"MeV\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The initial energy of each alpha-particle is 14.896 MeV\n"
- ]
- }
- ],
- "prompt_number": 33
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 32.10 Page no 875"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "M1=235.0439\n",
- "M2=93.9065\n",
- "M3=139.9055\n",
- "M4=1.00866\n",
- "S=931.5\n",
- "\n",
- "#Calculation\n",
- "Q=(M1-M2-M3-M4)*S\n",
- "\n",
- "#Result\n",
- "print\"Total energy is\",round(Q,2),\"MeV\"\n",
- "print\"All the available energy does not appear as the kinetic energy of fission products\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Total energy is 207.95 MeV\n",
- "All the available energy does not appear as the kinetic energy of fission products\n"
- ]
- }
- ],
- "prompt_number": 38
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 32.11 Page no 875"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "M1=2.014102\n",
- "M2=3.016049\n",
- "M3=4.002603\n",
- "M4=1.008665\n",
- "S=931.5\n",
- "q1=1.6*10**-19 #C\n",
- "q2=9*10**9\n",
- "r=2*1.5*10**-15\n",
- "r1=1.38*10**-23\n",
- "\n",
- "#Calculation\n",
- "Q=(M1+M2-M3-M4)*S\n",
- "Q2=q2*(q1**2/r)\n",
- "Q3=(2*Q2)/(3*r1)\n",
- "\n",
- "#Result\n",
- "print\"(a) The amount of energy is\",round(Q,2),\"MeV\"\n",
- "print\"(b) The kinetic energy is\",Q2,\"J\"\n",
- "print\"The temperature is\",round(Q3*10**-9,2),\"*10**9 K\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a) The amount of energy is 17.59 MeV\n",
- "(b) The kinetic energy is 7.68e-14 J\n",
- "The temperature is 3.71 *10**9 K\n"
- ]
- }
- ],
- "prompt_number": 55
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/modern_physics_by_Satish_K._Gupta/chap33.ipynb b/modern_physics_by_Satish_K._Gupta/chap33.ipynb deleted file mode 100755 index 4183f15e..00000000 --- a/modern_physics_by_Satish_K._Gupta/chap33.ipynb +++ /dev/null @@ -1,304 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:52e9ff37336a0b3392520c7ec0e817b37056adea4740de3297c85e4e3e83dc56"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 33 Solids"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 33.1 Page no 904"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "Eg=0.72*1.6*10**-19 #J\n",
- "h=6.62*10**-34\n",
- "c=3*10**8\n",
- "\n",
- "#Calculation\n",
- "L=(h*c)/Eg\n",
- "\n",
- "#Result\n",
- "print\"The maximum wavelength of electromagnetic radiation is\",round(L*10**6,3)*10**-6,\"m\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The maximum wavelength of electromagnetic radiation is 1.724e-06 m\n"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 33.2 Page no 904"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "n1=1.5*10**16 #/m**3\n",
- "nh=4.5*10**22\n",
- "\n",
- "#Calculation\n",
- "ne=n1**2/nh\n",
- "\n",
- "#Result\n",
- "print\"ne in the doping silicon is\",ne*10**-9,\"*10**9 /m**3\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "ne in the doping silicon is 5.0 *10**9 /m**3\n"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 33.3 Page no 904"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "ne=8*10**19 #/m**3\n",
- "nh=5*10**18\n",
- "ue=2.3 #m**2/V/S\n",
- "uh=0.01\n",
- "e=1.6*10**-19\n",
- "\n",
- "#Calculation\n",
- "a=1/(e*((ne*ue)+(nh*uh)))\n",
- "\n",
- "#Result\n",
- "print\"(a) The semiconductor has greater electron concentration, it is n-type semiconductor\"\n",
- "print\"(b) Resistivity is\", round(a*10**2,3),\"*10**-2 ohm/m\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a) The semiconductor has greater electron concentration, it is n-type semiconductor\n",
- "(b) Resistivity is 3.396 *10**-2 ohm/m\n"
- ]
- }
- ],
- "prompt_number": 16
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 33.4 Page no 904"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "e=1.6*10**-19 #C\n",
- "A=500 #ohm**-1 m**-1\n",
- "Ue=0.39 #m**2 V**-1 s**-1\n",
- "\n",
- "#Calculation\n",
- "Ne=A/(e*Ue)\n",
- "\n",
- "#Result\n",
- "print\"The number density of donor atoms is\",round(Ne*10**-21,3)*10**21,\"m**3\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The number density of donor atoms is 8.013e+21 m**3\n"
- ]
- }
- ],
- "prompt_number": 23
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 33.5 Page no 904"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "F=1.6*10**-19 #E\n",
- "W=4.2*10**8\n",
- "e=2.4\n",
- "w=4.2*10**-8\n",
- "\n",
- "#Calculation\n",
- "S=F*W\n",
- "A=S/F\n",
- "E=e/w\n",
- "\n",
- "#Result\n",
- "print\"Electric field is\", round(E*10**-7,2),\"*10**7 V/m\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Electric field is 5.71 *10**7 V/m\n"
- ]
- }
- ],
- "prompt_number": 30
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 33.6 Page no 904"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "h=6.62*10**-34\n",
- "c=3*10**8\n",
- "l=630*10**-9\n",
- "e=1.6*10**-19\n",
- "\n",
- "#Calculation\n",
- "Eg=(h*c)/(l*e)\n",
- "\n",
- "#Result\n",
- "print\"Width of the forbidden energy gap is\",round(Eg,2),\"eV\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Width of the forbidden energy gap is 1.97 eV\n"
- ]
- }
- ],
- "prompt_number": 33
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 33.7 Page no 904"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#iven\n",
- "A=10**-4 #m**2\n",
- "l=0.1 #m\n",
- "V=2 #V\n",
- "T=300 #K\n",
- "ue=0.135 #m**2/V/S\n",
- "n=1.5*10**15 #/m**3\n",
- "uh=0.048 #m**2/V/S\n",
- "e=1.6*10**-19\n",
- "ue1=0.39\n",
- "uh1=0.19\n",
- "n1=2.4*10**19\n",
- "\n",
- "#Calculation\n",
- "E=V/l\n",
- "ve=ue*E\n",
- "vh=uh*E\n",
- "Ie=e*A*n*ve\n",
- "Ih=e*A*n*vh\n",
- "I=Ie+Ih\n",
- "ve1=ue1*E\n",
- "ve2=uh1*E\n",
- "Ie1=e*A*n1*ve1\n",
- "Ie2=e*A*n1*ve2\n",
- "I1=Ie1+Ie2\n",
- "\n",
- "#Result\n",
- "print\"Electron current is\", Ie*10.0,\"A \\nHole current is\",Ih*10,\"A\"\n",
- "print\"Magnitude of total current is\",I*10,\"A \\nTotal current when germanium is used is\",I1*10**3,\"*10**-3 A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Electron current is 6.48e-07 A \n",
- "Hole current is 2.304e-07 A\n",
- "Magnitude of total current is 8.784e-07 A \n",
- "Total current when germanium is used is 4.4544 *10**-3 A\n"
- ]
- }
- ],
- "prompt_number": 48
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/modern_physics_by_Satish_K._Gupta/chap34.ipynb b/modern_physics_by_Satish_K._Gupta/chap34.ipynb deleted file mode 100755 index 9aea6687..00000000 --- a/modern_physics_by_Satish_K._Gupta/chap34.ipynb +++ /dev/null @@ -1,621 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:13c8f6846e205753ae8bd963e82228c66a5cc1ec104d777be242a6767f4586d5"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 34 Semiconductor devices"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 34.1 Page no 932"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "E=3 #V\n",
- "Vd=0.7\n",
- "R=100.0\n",
- "\n",
- "#Calculation\n",
- "V=E-Vd\n",
- "I=V/R\n",
- "\n",
- "#Result\n",
- "print\"Current in the circuit is\",I*10**3,\"mA\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Current in the circuit is 23.0 mA\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 34.2 Page no 932"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "E=1.5 #V\n",
- "Vd=0.5\n",
- "P=0.1 #W\n",
- "\n",
- "#Calculation\n",
- "I=P/Vd\n",
- "V=E-Vd\n",
- "R=V/I\n",
- "\n",
- "#Result\n",
- "print\"Value of resistance is\",R,\"ohm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Value of resistance is 5.0 ohm\n"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 34.3 Page no 932"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "E=9 #V\n",
- "Vz=6 #V\n",
- "Rl=1000.0 #ohm\n",
- "R=100.0 #ohm\n",
- "\n",
- "#Calculation\n",
- "V=E-Vz\n",
- "I=V/R\n",
- "Il=Vz/Rl\n",
- "Iz=I-Il\n",
- "Pz=Vz*Iz\n",
- "\n",
- "#Result\n",
- "print\"Power dissipated in zener diode is\",Pz,\"watt\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Power dissipated in zener diode is 0.144 watt\n"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 34.4 Page no 932"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "E0=50 #V\n",
- "S=2.0\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "E1=50/math.sqrt(S) #V\n",
- "V=math.sqrt(E1**2/S)\n",
- "V1=(S*E0/math.pi)/S\n",
- "\n",
- "#Result\n",
- "print\"(a) The r.m.s. voltage across Rl is\",V,\"V\"\n",
- "print\"(b) Reading of a d.c voltametre connected across Rl is\",round(V1,2),\"V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a) The r.m.s. voltage across Rl is 25.0 V\n",
- "(b) Reading of a d.c voltametre connected across Rl is 15.92 V\n"
- ]
- }
- ],
- "prompt_number": 22
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 34.5 Page no 933"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "A=0.96\n",
- "Ie=7.2 #mA\n",
- "\n",
- "#Calculation\n",
- "Ic=A*Ie\n",
- "Ib=Ie-Ic\n",
- "\n",
- "#Result\n",
- "print\"The base current is\",round(Ib,2),\"mA\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The base current is 0.29 mA\n"
- ]
- }
- ],
- "prompt_number": 27
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 34.6 Page no 933"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "B=70\n",
- "Ie=8.8 #mA\n",
- "Ib=1.0\n",
- "Ic=70.0\n",
- "\n",
- "#Calculation\n",
- "Ib1=Ie/(Ic+Ib)\n",
- "Ic1=Ic*Ib1\n",
- "A=Ic/(Ic+Ib)\n",
- "\n",
- "#Result\n",
- "print\"The collector current is\",round(Ib1,3),\"mA\"\n",
- "print\"The base current is\",round(Ic1,2),\"mA\"\n",
- "print\"Current gain is\",round(A,3)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The collector current is 0.124 mA\n",
- "The base current is 8.68 mA\n",
- "Current gain is 0.986\n"
- ]
- }
- ],
- "prompt_number": 12
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 34.7 Page no 933"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "Ib=105*10**-6 #A\n",
- "Ic=2.05*10**-3 #A\n",
- "Ib1=27*10**-6 #A\n",
- "Ic1=650*10**-6 #A\n",
- "\n",
- "#Calculation\n",
- "B=Ic/Ib\n",
- "Ie=Ib+Ic\n",
- "A=Ic/Ie\n",
- "Bac=Ic1/Ib1\n",
- "\n",
- "#Result\n",
- "print\"(a) The value of B is\",round(B,1),\",Value of Ie is\",Ie*10**3,\"*10**-3 A\",\"and Value of A is\",round(A,2)\n",
- "print\"(b) The value of Bac is\",round(Bac,2)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a) The value of B is 19.5 ,Value of Ie is 2.155 *10**-3 A and Value of A is 0.95\n",
- "(b) The value of Bac is 24.07\n"
- ]
- }
- ],
- "prompt_number": 29
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 34.8 Page no 933"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "Ie=7.89*10**-3 #A\n",
- "Ic=7.8*10**-3 #A\n",
- "\n",
- "#Calculation\n",
- "A=Ic/Ie\n",
- "B=A/(1-A)\n",
- "Ib=Ic/B\n",
- "\n",
- "#Result\n",
- "print\"change in the base current is\",Ib,\"A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "change in the base current is 9e-05 A\n"
- ]
- }
- ],
- "prompt_number": 39
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 34.9 Page no 933"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "Ib=20.0*10**-6 #A\n",
- "Vbe=0.02 #V\n",
- "Ic=2*10**-3 #A\n",
- "Rl=5000 #ohm\n",
- "\n",
- "#Calculation\n",
- "P=Vbe/Ib\n",
- "B=Ic/Ib\n",
- "Gm=Ic/Vbe\n",
- "Q=(Rl*Ic)/Vbe\n",
- "\n",
- "#Result\n",
- "print\"(a) The input resistance Bac is\",P,\"ohm\",\"and transconductance of the transister is\",Gm,\"ohm**-1\"\n",
- "print\"(b) The voltage gain of the amplifier is\",Q"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a) The input resistance Bac is 1000.0 ohm and transconductance of the transister is 0.1 ohm**-1\n",
- "(b) The voltage gain of the amplifier is 500.0\n"
- ]
- }
- ],
- "prompt_number": 47
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 34.10 Page no 933"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "Ib=15*10**-6 #A\n",
- "Ic=2*10**-3 #A\n",
- "R=665.0 #ohm\n",
- "Rl=5*10**3 #ohm\n",
- "\n",
- "#Calculation\n",
- "B=Ic/Ib\n",
- "Gm=B/R\n",
- "Av=Gm*Rl\n",
- "\n",
- "#Result \n",
- "print\"(i) Current gain is\",round(B,1)\n",
- "print\"(ii) Transconductance is\",round(Gm,1),\"ohm**-1\"\n",
- "print\"(iii) Voltage gain Av of the amplifier is\",round(Av,0)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(i) Current gain is 133.3\n",
- "(ii) Transconductance is 0.2 ohm**-1\n",
- "(iii) Voltage gain Av of the amplifier is 1003.0\n"
- ]
- }
- ],
- "prompt_number": 41
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 34.11 Page no 934"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "B=100.0\n",
- "Vcc=24 #V\n",
- "Ic=1.5*10**-3 #A\n",
- "Rc=4.7*10**3\n",
- "Rb=220*10**3\n",
- "\n",
- "#Calculation\n",
- "Ib=Ic/B\n",
- "Vce=Vcc-(Ic*Rc)\n",
- "Vbe=Vcc-(Ib*Rb)\n",
- "Vbc=(Ic*Rc)-(Ib*Rb)\n",
- "\n",
- "#Result\n",
- "print\"The value of Ib is\",Ib,\"A\"\n",
- "print\"The value of Vce is\",Vce,\"V\"\n",
- "print\"The value of Vbe is\",Vbe,\"V\"\n",
- "print\"The value of Vbc is\",Vbc,\"V\"\n",
- "print\"The transistor is in saturation state\" "
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of Ib is 1.5e-05 A\n",
- "The value of Vce is 16.95 V\n",
- "The value of Vbe is 20.7 V\n",
- "The value of Vbc is 3.75 V\n",
- "The transistor is in saturation state\n"
- ]
- }
- ],
- "prompt_number": 84
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 34.15 Page no 935"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "Vb=5 #V\n",
- "I=10**3 #A\n",
- "Q=0.7\n",
- "I1=5*10**-3\n",
- "Vb1=6 #V\n",
- "R0=1000 #ohm\n",
- "I3=10**-3 #A\n",
- "\n",
- "#Calculation\n",
- "R=(Vb-Q)*I\n",
- "R1=(R*10**-3)/I1\n",
- "R3=(Vb1-Q)/I1\n",
- "R4=I1**2*R3\n",
- "R5=Q*I1\n",
- "Vz=(I3*R0)+Q\n",
- "\n",
- "#Result\n",
- "print\"(a) The maximum value of R is\",R*10**-3,\"10**3\",\"ohm\"\n",
- "print\"(b) The value of R is\",R1,\"ohm\"\n",
- "print\"(c) The power dissipated across R is\",R4*10**3,\"10**-3\",\"ohm\",\"and across diode is\",R5*10**3,\"10**-3 W\"\n",
- "print\"(d) The maximum voltage Vb is\",Vz,\"V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a) The maximum value of R is 4.3 10**3 ohm\n",
- "(b) The value of R is 860.0 ohm\n",
- "(c) The power dissipated across R is 26.5 10**-3 ohm and across diode is 3.5 10**-3 W\n",
- "(d) The maximum voltage Vb is 1.7 V\n"
- ]
- }
- ],
- "prompt_number": 148
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 34.16 Page no 935"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "I0=5*10**-12 #A\n",
- "K=8.6*10**-5 #eVK**-1\n",
- "Q=1.6*10**-19 #J K**-1\n",
- "V=0.6 #volts\n",
- "V1=0.7 #volts\n",
- "A=23.256\n",
- "A1=27.132\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "I=-I0*(Q*V/math.exp(K*Q)-1)\n",
- "Z=I*(math.exp(A)-1)\n",
- "I1=-I0*(Q*V1/math.exp(K*Q)-1)\n",
- "Z1=I1*(math.exp(A1)-1)\n",
- "S=Z1-Z\n",
- "V2=V1-V\n",
- "J=V2/S\n",
- "\n",
- "#Result\n",
- "print\"(a) The forward current at a forward voltage is\",round(Z,4),\"A\"\n",
- "print\"(b) The voltage across the diode is\",round(S,4),\"A\"\n",
- "print\"(c) The dynamic resistance is\",round(J,4),\"ohm\"\n",
- "print\"(d) For change in votage from 1V to 2V,the current will remain equal to\",I0,\"A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a) The forward current at a forward voltage is 0.0629 A\n",
- "(b) The voltage across the diode is 2.9727 A\n",
- "(c) The dynamic resistance is 0.0336 ohm\n",
- "(d) For change in votage from 1V to 2V,the current will remain equal to 5e-12 A\n"
- ]
- }
- ],
- "prompt_number": 16
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 34.17 Page no 936"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "Vbe=24*10**-3 #V\n",
- "Ib=32*10**-6 #A\n",
- "Ic=3.6*10**-3 #A\n",
- "Rl=4.8*10**3\n",
- "\n",
- "#Calculation\n",
- "B=Ic/Ib\n",
- "Rbe=Vbe/Ib\n",
- "Gm=Ic/Vbe\n",
- "Av=(Rl*Ic)/Vbe\n",
- "\n",
- "#Result\n",
- "print\"(i) The current gain is\",B\n",
- "print\"(ii) The input resistance Rbe is\",Rbe,\"ohm\"\n",
- "print\"(iii) The transconductance Gm is\",Gm,\"S\"\n",
- "print\"(iv) Voltage gain Av is\",Av"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(i) The current gain is 112.5\n",
- "(ii) The input resistance Rbe is 750.0 ohm\n",
- "(iii) The transconductance Gm is 0.15 S\n",
- "(iv) Voltage gain Av is 720.0\n"
- ]
- }
- ],
- "prompt_number": 28
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/modern_physics_by_Satish_K._Gupta/chap35.ipynb b/modern_physics_by_Satish_K._Gupta/chap35.ipynb deleted file mode 100755 index 27056667..00000000 --- a/modern_physics_by_Satish_K._Gupta/chap35.ipynb +++ /dev/null @@ -1,263 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:6fd837c473d583739b16708ec5ff802d29867f6cb2f48bae07778822e1c04084"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 35 Analog and Digital Communication"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 35.1 Page no 978"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "Em=0.5 \n",
- "Ec=1\n",
- "\n",
- "#Calculation\n",
- "Emax=Ec+Em\n",
- "Emin=Ec-Em\n",
- "Ma=(Emax-Emin)/(Emax+Emin)\n",
- "\n",
- "#Result\n",
- "print\"The modulation index is\",Ma"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The modulation index is 0.5\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 35.2 Page no 978"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "L=3.6*10**-3 #H\n",
- "C=2.5*10**-12 #F\n",
- "Fm=15 #KHz\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "Fc=1/(2*math.pi*math.sqrt(L*C))*10**-3\n",
- "S=Fc+Fm\n",
- "S1=Fc-Fm\n",
- "\n",
- "#Result\n",
- "print\"Frequency of upper sideband is\",round(S,0),\"kHz\"\n",
- "print\"Frequency of lower sideband is\",round(S1,0),\"kHz\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Frequency of upper sideband is 1693.0 kHz\n",
- "Frequency of lower sideband is 1663.0 kHz\n"
- ]
- }
- ],
- "prompt_number": 27
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 35.3 Page no 978"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "Fm=3.2 #kHz\n",
- "Fc=84*10**3 #kHz\n",
- "A=96 #kHz\n",
- "\n",
- "#Calculation\n",
- "Mf=A/Fm\n",
- "M1=Fc-Fm\n",
- "M2=Fc+Fm\n",
- "\n",
- "#Result\n",
- "print\"(a) Frequency modulation index is\",Mf\n",
- "print\"(b) Frequency range of the modulated wave is\",round(M1*10**-3,3),\"*10**3 MHz\",\"to\",round(M2*10**-3,3),\"*10**3 MHz\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a) Frequency modulation index is 30.0\n",
- "(b) Frequency range of the modulated wave is 83.997 *10**3 MHz to 84.003 *10**3 MHz\n"
- ]
- }
- ],
- "prompt_number": 44
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 35.4 Page no 979"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "A=1\n",
- "B=2\n",
- "C=0\n",
- "\n",
- "#Calculation\n",
- "Q1=(A*B**0)+(C*B**1)+(C*B**2)+(C*B**3)\n",
- "Q2=(C*B**0)+(A*B**1)+(C*B**2)+(C*B**3)\n",
- "Q3=(C*B**0)+(C*B**1)+(A*B**2)+(C*B**3)\n",
- "Q4=(C*B**0)+(C*B**1)+(C*B**2)+(A*B**3)\n",
- "Q5=(A*B**0)+(C*B**1)+(C*B**2)+(A*B**3)\n",
- "Q6=(A*B**0)+(C*B**1)+(A*B**2)+(A*B**3)\n",
- "\n",
- "#Result\n",
- "print\"Input voltage 0001 is\",Q1,\"V\"\n",
- "print\"Input voltage 0010 is\",Q2,\"V\"\n",
- "print\"Input voltage 0100 is\",Q3,\"V\"\n",
- "print\"Input voltage 1000 is\",Q4,\"V\"\n",
- "print\"Input voltage 1001 is\",Q5,\"V\"\n",
- "print\"Input voltage 1101 is\",Q6,\"V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Input voltage 0001 is 1 V\n",
- "Input voltage 0010 is 2 V\n",
- "Input voltage 0100 is 4 V\n",
- "Input voltage 1000 is 8 V\n",
- "Input voltage 1001 is 9 V\n",
- "Input voltage 1101 is 13 V\n"
- ]
- }
- ],
- "prompt_number": 55
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 35.5 Page no 979"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "A=2\n",
- "N=32\n",
- "\n",
- "#Calculation\n",
- "Z=A**N\n",
- "\n",
- "#Result\n",
- "print\"Number of quantisation levels is\",round(Z*10**-9,1),\"*10**9\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Number of quantisation levels is 4.3 *10**9\n"
- ]
- }
- ],
- "prompt_number": 60
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 35.6 Page no 979"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "S=16\n",
- "Q=8000\n",
- "\n",
- "#Calculation\n",
- "N=math.sqrt(S)\n",
- "W=Q*N\n",
- "\n",
- "#Result\n",
- "print\"The bill rate for a signal is\",W,\"bill s**-1\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The bill rate for a signal is 32000.0 bill s**-1\n"
- ]
- }
- ],
- "prompt_number": 69
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/modern_physics_by_Satish_K._Gupta/chap36.ipynb b/modern_physics_by_Satish_K._Gupta/chap36.ipynb deleted file mode 100755 index 4e06b38b..00000000 --- a/modern_physics_by_Satish_K._Gupta/chap36.ipynb +++ /dev/null @@ -1,103 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:d9296ed3bbb0144b6fddcd9bd9b4d00b56d458558332cd059b27e6750c38d570"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 36 Space Communication"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 36.1 Page no 990"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "AB=2*10**5 #m\n",
- "CN=100 #km\n",
- "C=3.0*10**8\n",
- "\n",
- "#Calculation\n",
- "Tg=AB/C\n",
- "Q=(math.sqrt(CN**2+CN**2))+(math.sqrt(CN**2+CN**2))\n",
- "Ts=Q*10**3/C\n",
- "W=(Ts-Tg)\n",
- "\n",
- "#Result\n",
- "print\"The time delay is\",round(W*10**4,2),\"*10**-4 s\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The time delay is 2.76 *10**-4 s\n"
- ]
- }
- ],
- "prompt_number": 57
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 36.2 Page no 990"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "H=80 #m\n",
- "R=6.4*10**6\n",
- "W=800 #km**-2\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "D=math.sqrt(2*H*R)\n",
- "A=math.pi*D**2\n",
- "S=A*W\n",
- "\n",
- "#Result\n",
- "print\"(a) The radius of the circle is\",D*10**-4,\"10**4 m\"\n",
- "print\"(b) The population covered by the transmission is\",round(S*10**-12,3),\"*10**6\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a) The radius of the circle is 3.2 10**4 m\n",
- "(b) The population covered by the transmission is 2.574 *10**6\n"
- ]
- }
- ],
- "prompt_number": 32
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/modern_physics_by_Satish_K._Gupta/chap37.ipynb b/modern_physics_by_Satish_K._Gupta/chap37.ipynb deleted file mode 100755 index 03f26ef4..00000000 --- a/modern_physics_by_Satish_K._Gupta/chap37.ipynb +++ /dev/null @@ -1,183 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:10e3d0cfbf86d2f1081a9d348d19bd93a499331702a8a9f3c32ddc1a8679292a"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 37 Line Communication"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 37.1 Page no 1007"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "U=1.48\n",
- "U2=1.45\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "C=U2/U\n",
- "C1=math.asin(C)*180/3.14\n",
- "\n",
- "#Result \n",
- "print\"The critical angle for a light is\",round(C1,1),\"degree\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The critical angle for a light is 78.5 degree\n"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 37.2 Page no 1007"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "u=1.7\n",
- "u2=1.5\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "C=u2/u\n",
- "A=math.asin(C)*180/3.14\n",
- "I=math.sqrt(u**2-u2**2)\n",
- "A1=math.asin(I)*180/3.14\n",
- "\n",
- "#Result\n",
- "print\"The critical angle is\",round(A,1),\"degree\"\n",
- "print\"The angle of acceptance is\",round(A1,1),\"degree\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The critical angle is 62.0 degree\n",
- "The angle of acceptance is 53.2 degree\n"
- ]
- }
- ],
- "prompt_number": 19
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 37.3 Page no 1007"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "u1=1.5\n",
- "u2=1.3\n",
- "L=30 #m\n",
- "c=3*10**8\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "V=c/u1\n",
- "t1=L/V\n",
- "C=u2/u1\n",
- "C1=math.asin(C)*180/3.14\n",
- "x=L/C\n",
- "t=x/V\n",
- "\n",
- "#Result\n",
- "print\"(a) Time taken for light in axical mode is\",t1*10**7,\"*10**-7 s\" \n",
- "print\"(b) Time taken for light in highest order mode is\",round(t*10**7,2),\"*10**-7 s\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a) Time taken for light in axical mode is 1.5 *10**-7 s\n",
- "(b) Time taken for light in highest order mode is 1.73 *10**-7 s\n"
- ]
- }
- ],
- "prompt_number": 49
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 37.5 Page no 1007"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "a=0.5\n",
- "b=50.0 #Km\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "A=10*math.log10(a)\n",
- "B=A/b\n",
- "\n",
- "#Result\n",
- "print\"Attenuation is\", round(B,2),\"dB/Km\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Attenuation is -0.06 dB/Km\n"
- ]
- }
- ],
- "prompt_number": 54
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/modern_physics_by_Satish_K._Gupta/chap4.ipynb b/modern_physics_by_Satish_K._Gupta/chap4.ipynb deleted file mode 100755 index 488631fb..00000000 --- a/modern_physics_by_Satish_K._Gupta/chap4.ipynb +++ /dev/null @@ -1,397 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:1087fb8dbcb5b183b8a1d36b431412b33d35209d5f6a3a237469594e817213fb"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 4 Gauss theorem"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 4.1 Page no 88"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "E=25 #V/m\n",
- "s=150*10**-4 #m**2\n",
- "a=60 #degree\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "A=E*s*math.cos(a*3.14/180.0)\n",
- "\n",
- "#Result\n",
- "print\"Flux of the electric field is\",round(A,4),\"Nm**2/C\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Flux of the electric field is 0.1877 Nm**2/C\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 4.2 Page no 88"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "E=3*10**3 #N/C\n",
- "S=10**-2 #m**2\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "a=E*S\n",
- "A=E*S*math.cos(60*3.14/180.0)\n",
- "\n",
- "#Result\n",
- "print\"(a) Flux of the field is\",a,\"Nm**2/C\"\n",
- "print\"(b) Flux through the square is\", round(A,0),\"Nm**2/C\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a) Flux of the field is 30.0 Nm**2/C\n",
- "(b) Flux through the square is 15.0 Nm**2/C\n"
- ]
- }
- ],
- "prompt_number": 10
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 4.3 Page no 88"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "e=8.854*10**-12 #C**2/N/m**2\n",
- "q=1 #C\n",
- "\n",
- "#Calculation\n",
- "a=q/e\n",
- "\n",
- "#Result\n",
- "print\"Number of electric lines are\",round(a*10**-11,3)*10**11"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Number of electric lines are 1.129e+11\n"
- ]
- }
- ],
- "prompt_number": 14
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 4.4 Page no 88"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "a=8*10**3 #Nm**2/C\n",
- "e=8.854*10**-12 #C**2/N/m**2\n",
- "\n",
- "#Calculation\n",
- "q=a*e\n",
- "\n",
- "#Result\n",
- "print\"(a) Net charge inside the box is\",q,\"C\"\n",
- "print\"(b) If the net outward flux is zero,we can't conclude that the charge inside the box is zero.\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a) Net charge inside the box is 7.0832e-08 C\n",
- "(b) If the net outward flux is zero,we can't conclude that the charge inside the box is zero.\n"
- ]
- }
- ],
- "prompt_number": 18
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 4.5 Page no 89"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "q=10**-5\n",
- "e=8.854*10**-12\n",
- "\n",
- "#Calculation\n",
- "a=q/(6.0*e)\n",
- "\n",
- "#Result\n",
- "print\"Magnitude of electric flux is\", round(a*10**-5,2),\"*10**5 Nm**2/C\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Magnitude of electric flux is 1.88 *10**5 Nm**2/C\n"
- ]
- }
- ],
- "prompt_number": 23
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 4.6 Page no 89"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "E=9*10**4 #N/C\n",
- "r=0.04 #m\n",
- "m=9*10**9\n",
- "\n",
- "#Calculation\n",
- "l=E*r/(2.0*m)\n",
- "\n",
- "#Result\n",
- "print\"Linear charge density is\", l,\"C/m\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Linear charge density is 2e-07 C/m\n"
- ]
- }
- ],
- "prompt_number": 26
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 4.7 Page no 89"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "a=5*10**-6 #C/m**2\n",
- "e=8.854*10**-12 #C**2/Nm**2\n",
- "r=0.1 #m\n",
- "a1=60 #Degree\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "A=(a*math.pi*r**2*math.cos(a1*3.14/180.0))/(2.0*e)\n",
- "\n",
- "#Result\n",
- "print\"Electric flux is\", round(A*10**-3,2),\"*10**3 N m**2/C\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Electric flux is 4.44 *10**3 N m**2/C\n"
- ]
- }
- ],
- "prompt_number": 31
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 4.8 Page no 89"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "a=17*10**-22 #C/m**2\n",
- "e=8.854*10**-12\n",
- "\n",
- "#Calculation\n",
- "E=a/e\n",
- "\n",
- "#Result\n",
- "print\"(a) Electric field to the left of the plate is zero\"\n",
- "print\"(b) Electric field to the right of the plate is zero\"\n",
- "print\"(c) Electric field between the plates is\",round(E*10**10,2)*10**-10,\"N/C\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a) Electric field to the left of the plate is zero\n",
- "(b) Electric field to the right of the plate is zero\n",
- "(c) Electric field between the plates is 1.92e-10 N/C\n"
- ]
- }
- ],
- "prompt_number": 38
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 4.9 Page no 89"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "q=1.6*10**-7 #C\n",
- "R=0.12 #m\n",
- "m=9*10**9\n",
- "r=0.18\n",
- "\n",
- "#Calculation\n",
- "E=(m*q)/R**2\n",
- "E1=(m*q)/r**2\n",
- "\n",
- "#Result\n",
- "print\"(a) ELectric field inside the sphere is zero\"\n",
- "print\"(b) Electric field outside the sphere is\",E,\"N/C\"\n",
- "print\"(c) Electric field at a point 18 cm from the centre is\", round(E1*10**-4,2),\"*10**4 N/C\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a) ELectric field inside the sphere is zero\n",
- "(b) Electric field outside the sphere is 100000.0 N/C\n",
- "(c) Electric field at a point 18 cm from the centre is 4.44 *10**4 N/C\n"
- ]
- }
- ],
- "prompt_number": 51
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 4.11 Page no 90"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "e=50\n",
- "V=0.2\n",
- "m=9*10**9\n",
- "\n",
- "#Calculation\n",
- "q=e*V**2/m\n",
- "\n",
- "#Result\n",
- "print\"Charge contained in the sphere is\",round(q*10**10,2)*10**-10,\"C\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Charge contained in the sphere is 2.22e-10 C\n"
- ]
- }
- ],
- "prompt_number": 56
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/modern_physics_by_Satish_K._Gupta/chap5.ipynb b/modern_physics_by_Satish_K._Gupta/chap5.ipynb deleted file mode 100755 index 84950cb9..00000000 --- a/modern_physics_by_Satish_K._Gupta/chap5.ipynb +++ /dev/null @@ -1,1167 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:59a10421d0b18543bcaaf15fad8dcc2fa1ef98c5fe857c6feab5eddfb8fae6cd"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 5 Capacitor"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 5.1 Page no 108"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "r=6.4*10**6 #m\n",
- "t=9*10**9\n",
- "\n",
- "#Calculation\n",
- "C=r/t\n",
- "\n",
- "#Result\n",
- "print\"The capacitance is\",round(C*10**6,0),\"F\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The capacitance is 711.0 F\n"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "Example 5.2 Page no 108"
- ],
- "language": "python",
- "metadata": {},
- "outputs": []
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "r=0.01 #m\n",
- "q=1 #C\n",
- "d=9.0*10**9\n",
- "\n",
- "#Calculation\n",
- "V=(q*d)/r\n",
- "\n",
- "#Result\n",
- "print\"V=\",V,\"V\"\n",
- "print\"The given metal sphere will not be able to hold charge of 1 C\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "V= 9e+11 V\n",
- "The given metal sphere will not be able to hold charge of 1 C\n"
- ]
- }
- ],
- "prompt_number": 25
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 5.3 Page no 109"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "R=3\n",
- "Q=27\n",
- "q=4*220\n",
- "\n",
- "#Calculation\n",
- "d=(Q*q)/(4*R)\n",
- "\n",
- "#Result\n",
- "print\"The potential of the bigger drop is\",d,\"V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The potential of the bigger drop is 1980 V\n"
- ]
- }
- ],
- "prompt_number": 28
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "Example 5.4 Page no 109"
- ],
- "language": "python",
- "metadata": {},
- "outputs": []
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "C=2 #F\n",
- "d=0.5*10**-2 #m\n",
- "a=8.854*10**-12\n",
- "\n",
- "#Calculation\n",
- "Q=(C*d)/a\n",
- "\n",
- "#Result\n",
- "print\"Area of the plate is\", round(Q*10**-9,2),\"*10**9 m**2\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Area of the plate is 1.13 *10**9 m**2\n"
- ]
- }
- ],
- "prompt_number": 36
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 5.5 Page no 109"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "A=10**-2 #m**2\n",
- "d=10**-3 #m\n",
- "q=0.12*10**-6\n",
- "V=120.0\n",
- "e=8.85*10**-12\n",
- "\n",
- "#Calculation\n",
- "C=q/V\n",
- "K=C*d/(e*A)\n",
- "\n",
- "#Result\n",
- "print\"Dielectric constant of the material is\", round(K,1)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Dielectric constant of the material is 11.3\n"
- ]
- }
- ],
- "prompt_number": 40
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 5.6 Page no 109"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "A1=100.0 #cm**2\n",
- "A2=500 #cm**2\n",
- "d1=0.05 #cm\n",
- "\n",
- "#Calculation\n",
- "d2=(A2*d1)/A1\n",
- "\n",
- "#Result\n",
- "print\"Difference between the plates of second capacitor is\", d2,\"cm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Difference between the plates of second capacitor is 0.25 cm\n"
- ]
- }
- ],
- "prompt_number": 43
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 5.7 Page no 109"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "a=6.45*10**6 #m\n",
- "b=6.4*10**6 #m\n",
- "m=9*10**9\n",
- "\n",
- "#Calculation\n",
- "C=(a*b)/(m*(a-b))\n",
- "\n",
- "#Result\n",
- "print\"Capacitance of the spherical capacitor is\",round(C,3),\"F\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Capacitance of the spherical capacitor is 0.092 F\n"
- ]
- }
- ],
- "prompt_number": 48
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 5.8 Page no 109"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "a=12*10**-2 #m\n",
- "b=13*10**-2 #m\n",
- "q=2.5*10**-6 #C\n",
- "K=32\n",
- "m=9.0*10**9\n",
- "\n",
- "#Calculation\n",
- "C=(K*a*b)/(m*(b-a))\n",
- "V=q/C\n",
- "C1=a/m\n",
- "C2=C/C1\n",
- "\n",
- "#Result\n",
- "print\"(a) Capacitance of the capacitor is\", round(C*10**9,3)*10**-9,\"F\"\n",
- "print\"(b) Potential of the inner sphere is\", round(V,1),\"V\"\n",
- "print\"(c) Capacitance of the capacitor is\",C2"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a) Capacitance of the capacitor is 5.547e-09 F\n",
- "(b) Potential of the inner sphere is 450.7 V\n",
- "(c) Capacitance of the capacitor is 416.0\n"
- ]
- }
- ],
- "prompt_number": 61
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 5.9 Page no 110"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "a=1.4*10**-2 #m\n",
- "b=1.5*10**-2 #m\n",
- "q=3.5*10**-6 #C\n",
- "e=8.854*10**-12\n",
- "l=15*10**-2\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "C=(2*math.pi*e*l)/(2.303*math.log10(b/a))\n",
- "V=q/C\n",
- "\n",
- "#Result\n",
- "print\"Capacitance of the system is\", round(C*10**10,2)*10**-8,\"F\"\n",
- "print\"Potential of the inner cylinder is\", round(V*10**-4,2),\"*10**4 V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Capacitance of the system is 1.21e-08 F\n",
- "Potential of the inner cylinder is 2.89 *10**4 V\n"
- ]
- }
- ],
- "prompt_number": 10
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 5.10 Page no 110"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "C=9.0*10**-12 #F\n",
- "V=120 #V\n",
- "\n",
- "#Calculation\n",
- "C1=1/(3/C)\n",
- "q=(V*C)/3.0\n",
- "V1=q/C\n",
- "\n",
- "#Result\n",
- "print\"(a) Total capacitance is\",C1,\"F\"\n",
- "print\"(b) Potential difference across each capacitor is\",V1,\"V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a) Total capacitance is 3e-12 F\n",
- "(b) Potential difference across each capacitor is 40.0 V\n"
- ]
- }
- ],
- "prompt_number": 17
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 5.11 Page no 110"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "C1=2*10**-12 #F\n",
- "C2=3*10**-12\n",
- "C3=4*10**-12\n",
- "V=100 #Volts\n",
- "\n",
- "#Calculation\n",
- "C=C1+C2+C3\n",
- "q1=C1*V\n",
- "q2=C2*V\n",
- "q3=C3*V\n",
- "\n",
- "#Result\n",
- "print\"(a) Total capacitamce of the parallel combination is\",C,\"F\"\n",
- "print\"(b) Charge on q1 is\",q1,\"C\"\n",
- "print\"(c) Charge on q2 is\",q2,\"C\"\n",
- "print\"(d) Charge on q3 is\",q3,\"C\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a) Total capacitamce of the parallel combination is 9e-12 F\n",
- "(b) Charge on q1 is 2e-10 C\n",
- "(c) Charge on q2 is 3e-10 C\n",
- "(d) Charge on q3 is 4e-10 C\n"
- ]
- }
- ],
- "prompt_number": 26
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 5.12 Page no 110"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "c1=5.0 #micro F\n",
- "c2=4.0\n",
- "c3=3.0\n",
- "\n",
- "#Calculation\n",
- "C=1/(1/c1+1/c2)\n",
- "c11=C+c3\n",
- "\n",
- "#Result\n",
- "print\"Capacitance of the combination is\",round(c11,2),\"micro F\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Capacitance of the combination is 5.22 micro F\n"
- ]
- }
- ],
- "prompt_number": 31
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 5.13 Page no 111"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "cs=3 #micro F\n",
- "cp=16\n",
- "c1=12\n",
- "\n",
- "#Calculation\n",
- "C=cs*cp\n",
- "c2=cp-c1\n",
- "\n",
- "#Result\n",
- "print\"Capacitance of each capacitor is\",c2,\"micro F\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Capacitance of each capacitor is 4 micro F\n"
- ]
- }
- ],
- "prompt_number": 35
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 5.14 Page no 111"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "c2=6.0 #micro F\n",
- "c3=6.0\n",
- "c1=12.0\n",
- "c4=12.0\n",
- "\n",
- "#Calculation\n",
- "c23=c2+c3\n",
- "C=1/(1/c1+1/c23+1/c4)\n",
- "\n",
- "#Result\n",
- "print\"resultant of capacitance is\",C,\"micro F\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "resultant of capacitance is 4.0 micro F\n"
- ]
- }
- ],
- "prompt_number": 39
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 5.15 Page no 111"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "c1=3.0 #pico F\n",
- "c2=2.0\n",
- "\n",
- "#Calculation\n",
- "C=1/(1/c1+1/c2+1/c1)\n",
- "c11=C+c2\n",
- "C11=1/(1/c1+1/c11+1/c1)\n",
- "\n",
- "#Result \n",
- "print\"Resultant capacitance is\",round(C11,0),\"pF\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Resultant capacitance is 1.0 pF\n"
- ]
- }
- ],
- "prompt_number": 43
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 5.16 Page no 111"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "C2=2\n",
- "C3=2\n",
- "C1=1 #micro F\n",
- "C4=2\n",
- "C5=1\n",
- "\n",
- "#Calculation\n",
- "C23=C2*C3/(C2+C3)\n",
- "C123=C1+C23\n",
- "C1234=C123*C4/(C123+C4)\n",
- "C=C1234+C5\n",
- "\n",
- "#Ressult\n",
- "print\"Equivalent capacitance is\", C,\"micro F\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Equivalent capacitance is 2 micro F\n"
- ]
- }
- ],
- "prompt_number": 49
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 5.17 Page no 112"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "C5=10*10**-6 #micro F\n",
- "C6=10*10**-6\n",
- "C2=60.0*10**-6\n",
- "C3=60.0*10**-6\n",
- "C4=60.0*10**-6\n",
- "C1=40*10**-6\n",
- "V=100 #V\n",
- "\n",
- "#Calculation\n",
- "C56=(C5+C6)\n",
- "C234=1/(1/C2+1/C3+1/C4)\n",
- "C=C56+C234\n",
- "C11=(C1*C)/(C1+C)\n",
- "q=C56*V\n",
- "\n",
- "#Result\n",
- "print\"Total charge is\",q*10**3,\"*10**-3 C\"\n",
- "print\"Equivalent capacitance is\",C11*10**6,\"micro F\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Total charge is 2.0 *10**-3 C\n",
- "Equivalent capacitance is 20.0 micro F\n"
- ]
- }
- ],
- "prompt_number": 75
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 5.19 Page no 113"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "C1=5.0*10**-6 #F\n",
- "C2=10.0*10**-6\n",
- "C3=2.0*10**-6\n",
- "C4=4.0*10**-6\n",
- "\n",
- "#Calculation\n",
- "C12=1/(1/C1+1/C2)\n",
- "C34=1/(1/C3+1/C4)\n",
- "C=C12+C34\n",
- "\n",
- "#Result\n",
- "print\"Effective capacitance is\", round(C*10**6,2),\"micro F\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Effective capacitance is 4.67 micro F\n"
- ]
- }
- ],
- "prompt_number": 81
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 5.20 Page no 113"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "n=3\n",
- "C1=1.0\n",
- "C2=1.0\n",
- "C3=1.0\n",
- "C=2 #micro F\n",
- "\n",
- "#Calculation\n",
- "C11=1/(1/C1+1/C2+1/C3)\n",
- "m=C/C11\n",
- "\n",
- "#Result\n",
- "print\"Capacitance required is\", m"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Capacitance required is 6.0\n"
- ]
- }
- ],
- "prompt_number": 84
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 5.21 Page no 113"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "C1=4\n",
- "V=6\n",
- "C2=6\n",
- "\n",
- "#Calculation\n",
- "q1=C1*V\n",
- "q2=C2*V\n",
- "\n",
- "#Result\n",
- "print\"Charge on C1 is\",q1,\"micro C\"\n",
- "print\"Charge on C2 is\",q2,\"micro C\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Charge on C1 is 24 micro C\n",
- "Charge on C2 is 36 micro C\n"
- ]
- }
- ],
- "prompt_number": 88
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 5.22 Page no 114"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "C2=5 #micro F\n",
- "C3=5 \n",
- "a=6\n",
- "C1=10\n",
- "V=3\n",
- "\n",
- "#Calculation\n",
- "C23=C2+C3\n",
- "V1=a/2.0\n",
- "q1=C1*V\n",
- "q23=C23*V\n",
- "q3=q23/2.0\n",
- "\n",
- "#Result\n",
- "print\"Charge on capacitor C1 is\",q1,\"micro F\"\n",
- "print\"Charge on capacitor C2 is\",q23,\"micro F\"\n",
- "print\"Charge on capacitor C3 is\",q3,\"micro F\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Charge on capacitor C1 is 30 micro F\n",
- "Charge on capacitor C2 is 30 micro F\n",
- "Charge on capacitor C3 is 15.0 micro F\n"
- ]
- }
- ],
- "prompt_number": 94
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 5.23 Page no 114"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "C1=4*10**-6 #F\n",
- "V1=400 #V\n",
- "C2=2*10**-6\n",
- "q2=0\n",
- "\n",
- "#Calculation\n",
- "q1=C1*V1\n",
- "C=C1+C2\n",
- "q=q1+q2\n",
- "V=q/C\n",
- "\n",
- "#Result\n",
- "print\"Common potential is\", round(V,2),\"V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Common potential is 266.67 V\n"
- ]
- }
- ],
- "prompt_number": 101
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 5.24 Page no 114"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "V=12\n",
- "C1=8 #micro F\n",
- "C2=4\n",
- "\n",
- "#Calculation\n",
- "Va=V-C2\n",
- "Vb=V-C1\n",
- "V1=Va-Vb\n",
- "\n",
- "#Result\n",
- "print\"Potential difference is\", V1,\"V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Potential difference is 4 V\n"
- ]
- }
- ],
- "prompt_number": 105
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 5.25 Page no 114"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "A=90*10**-4 #m**2\n",
- "d=2.5*10**-3 #m\n",
- "V=400 #Volts\n",
- "e=8.854*10**-12\n",
- "\n",
- "#Calculation\n",
- "C=(e*A)/d\n",
- "W=(C*V**2)/2.0\n",
- "\n",
- "#Result\n",
- "print\"Electrostatic energy is\", round(W*10**6,2)*10**-6,\"J\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Electrostatic energy is 2.55e-06 J\n"
- ]
- }
- ],
- "prompt_number": 112
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 5.26 Page no 114"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "C=2000*10**-6 \n",
- "V=1.5 #V\n",
- "t=10.0**-4 #s\n",
- "\n",
- "#Calculation\n",
- "U=(C*V**2)/2.0\n",
- "P=U/t\n",
- "\n",
- "#Result\n",
- "print\"Energy stored in the capacitor is\", U*10**3,\"*10**-3 J\"\n",
- "print\"Power of the flash is\",P,\"W\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Energy stored in the capacitor is 2.25 *10**-3 J\n",
- "Power of the flash is 22.5 W\n"
- ]
- }
- ],
- "prompt_number": 119
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 5.27 Page no 115"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "C1=4*10**-6\n",
- "V1=200 #V\n",
- "C2=2*10**-6 # F\n",
- "\n",
- "#Calculation\n",
- "U1=(C1*V1**2)/2.0\n",
- "q=C1*V1\n",
- "C=C1+C2\n",
- "V=q/C\n",
- "U2=(C*V**2)/2.0\n",
- "U=U1-U2\n",
- "\n",
- "#Result\n",
- "print\"Electrostatic energy is\",round(U*10**2,2),\"*10**-2 J\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Electrostatic energy is 2.67 *10**-2 J\n"
- ]
- }
- ],
- "prompt_number": 129
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 5.28 Page no 115"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "K=6.0\n",
- "C=17.7*10**-12 #F\n",
- "V=100\n",
- "\n",
- "#Calculation\n",
- "C1=K*C\n",
- "q=C1*V\n",
- "V1=V/K\n",
- "q1=C1*V1\n",
- "\n",
- "#Result\n",
- "print\"(a) Charge on capacitor is\", q,\"C\"\n",
- "print\"(b) Charge on capacitor is\", q1,\"C\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a) Charge on capacitor is 1.062e-08 C\n",
- "(b) Charge on capacitor is 1.77e-09 C\n"
- ]
- }
- ],
- "prompt_number": 135
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 5.29 Page no 115"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "K=3\n",
- "e=8.854*10**-12 #C**2/N/m**2\n",
- "C=50*10**-12 #F\n",
- "a=10**7 #V/m\n",
- "b=10\n",
- "V=10**3\n",
- "\n",
- "#Calculation\n",
- "E=(a*b)/100.0\n",
- "q=C*V\n",
- "A=q/(e*K*E)\n",
- "\n",
- "#Result\n",
- "print\"Minimum area is\", round(A*10**4,1),\"cm**2\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Minimum area is 18.8 cm**2\n"
- ]
- }
- ],
- "prompt_number": 140
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 5.30 Page no 115"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "V=15*10**5 #V\n",
- "E=5.0*10**7 #V/m\n",
- "\n",
- "#Calculation\n",
- "r=V/E\n",
- "\n",
- "#Result\n",
- "print\"Minimum radius of the spherical shell is\", r*10**2,\"cm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Minimum radius of the spherical shell is 3.0 cm\n"
- ]
- }
- ],
- "prompt_number": 144
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/modern_physics_by_Satish_K._Gupta/chap6.ipynb b/modern_physics_by_Satish_K._Gupta/chap6.ipynb deleted file mode 100755 index 2c3272c0..00000000 --- a/modern_physics_by_Satish_K._Gupta/chap6.ipynb +++ /dev/null @@ -1,1695 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:9ba957b16289f27dba06e59c72667002157ed6d6f12727710ff725257f8649eb"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 6 Current electricity"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 6.1 Page no 191"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "I=300*10**-3\n",
- "e=1.6*10**-19 #C\n",
- "t=60 #S\n",
- "\n",
- "#Calculation\n",
- "q=I*t\n",
- "n=q/e\n",
- "\n",
- "#Result\n",
- "print\"Number of electron passed is\", n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Number of electron passed is 1.125e+20\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 6.2 Page no 191"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "e=1.6*10**-19 #C\n",
- "v=6.8*10**15 #revolution/s\n",
- "r=0.51*10**-10 #m\n",
- "\n",
- "#Calculation\n",
- "I=e*v\n",
- "\n",
- "#Result\n",
- "print\"Equivalent current is\", I*10**3,\"*10**-3 A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Equivalent current is 1.088 *10**-3 A\n"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 6.3 Page no 191 "
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "I=10 #A\n",
- "n=9*10**28 #/m**3\n",
- "A=10.0**-4 #m**2\n",
- "e=1.6*10**-19\n",
- "\n",
- "#Calculation\n",
- "Vd=I/(n*e*A)\n",
- "\n",
- "#Result\n",
- "print\"Drift velocity is\", round(Vd*10**6,2)*10**-6,\"m/s\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Drift velocity is 6.94e-06 m/s\n"
- ]
- }
- ],
- "prompt_number": 13
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 6.4 Page no 191"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "n=8.5*10**28 #/m**3\n",
- "I=3 #A\n",
- "A=2.0*10**-6 #m**2\n",
- "l=3 #m\n",
- "e=1.6*10**-19\n",
- "\n",
- "#Calculation\n",
- "Vd=I/(n*e*A)\n",
- "t=I/Vd\n",
- "\n",
- "#Result\n",
- "print\"Drift velocity of the electron is\", round(Vd*10**4,3),\"*10**-4 m/s\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Drift velocity of the electron is 1.103 *10**-4 m/s\n"
- ]
- }
- ],
- "prompt_number": 22
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 6.5 Page no 191"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "A=10**-7 #m**2\n",
- "I=1 #A\n",
- "K=6.023*10**23\n",
- "w=63.5\n",
- "d=9*10**3 #Kg/m**3\n",
- "\n",
- "#Calculation\n",
- "a=(K/w)*1000\n",
- "n=a*d\n",
- "Vd=I/(n*e*A)\n",
- "\n",
- "#Result\n",
- "print\"Average drift velocity is\", round(Vd*10**4,2),\"*10**-4 m/s\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Average drift velocity is 7.32 *10**-4 m/s\n"
- ]
- }
- ],
- "prompt_number": 30
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 6.6 Page no 191"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "V=5 #Volt\n",
- "l=0.1 #m\n",
- "vd=2.5*10**-4 #m/s\n",
- "\n",
- "#Calculation\n",
- "E=V/l\n",
- "u=vd/E\n",
- "\n",
- "#Result\n",
- "print\"Electron mobility is\", u,\"m**2/V/s\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Electron mobility is 5e-06 m**2/V/s\n"
- ]
- }
- ],
- "prompt_number": 34
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 6.7 Page no 191"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "V=3 #Volt\n",
- "l=0.2 #m\n",
- "A=10**-6 #m**2\n",
- "n=8.5*10**28 #/m**3\n",
- "u=4.5*10**-6 #m**2/V/s\n",
- "e=1.6*10**-19\n",
- "\n",
- "#Calculation\n",
- "E=V/l\n",
- "I=n*A*u*E*e\n",
- "\n",
- "#Result\n",
- "print\"Current through the wire is\",round(I,2),\"A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Current through the wire is 0.92 A\n"
- ]
- }
- ],
- "prompt_number": 37
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 6.8 Page no 191"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "I=1.8 #A\n",
- "A=0.5*10**-6 #m**2\n",
- "n=8.8*10**28\n",
- "e=1.6*10**-19\n",
- "\n",
- "#Calculation\n",
- "j=I/A\n",
- "vd=j/(n*e)\n",
- "\n",
- "#Result\n",
- "print\"Drift velocity is\", round(vd*10**4,2),\"*10**-4 m/s\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Drift velocity is 2.56 *10**-4 m/s\n"
- ]
- }
- ],
- "prompt_number": 45
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 6.9 Page no 192"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "V=1.5 #V\n",
- "n=8.4*10**28 #/m**3\n",
- "l=0.2 #m\n",
- "A=0.3*10**-6 #m**2\n",
- "I=2.4 #A\n",
- "m=9.1*10**-31 #Kg\n",
- "e=1.6*10**-19\n",
- "\n",
- "#Calculation\n",
- "E=V/l\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": 53
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 6.10 Page no 192"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "V=200 #V\n",
- "R=100.0 #ohm\n",
- "t=1\n",
- "\n",
- "#Calculation\n",
- "I=V/R\n",
- "q=I*t\n",
- "n=q/e\n",
- "\n",
- "#Result\n",
- "print\"Number of electrons is\", n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Number of electrons is 1.25e+19\n"
- ]
- }
- ],
- "prompt_number": 55
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 6.11 Page no 192"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "R=2 #ohm\n",
- "l=1 #m\n",
- "d=4*10**-4 #m\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "A=(math.pi*d**2)/4.0\n",
- "a=(R*A)/l\n",
- "\n",
- "#Result\n",
- "print\"Resistivity of the material is\", round(a*10**7,3)*10**-7,\"ohm m\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Resistivity of the material is 2.513e-07 ohm m\n"
- ]
- }
- ],
- "prompt_number": 62
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 6.12 Page no 192"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "m=0.45 #Kg\n",
- "R=0.14 #ohm\n",
- "a=1.78*10**-8 #ohm m\n",
- "l=1.99\n",
- "d=8.93*10**3\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "l=(math.sqrt(R*m/(a*d)))/10.0\n",
- "r=math.sqrt(m/(math.pi*l*d))\n",
- "\n",
- "#Result\n",
- "print\"Radius is\", round(r*10**3,2),\"mm\"\n",
- "print\"Length is\", round(l,2),\"m\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Radius is 2.84 mm\n",
- "Length is 1.99 m\n"
- ]
- }
- ],
- "prompt_number": 76
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 6.13 Page no 192"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "R=5\n",
- "\n",
- "#Calculation\n",
- "R1=4*R\n",
- "\n",
- "#Result\n",
- "print\"New resistance is\",R1,\"ohm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "New resistance is 20 ohm\n"
- ]
- }
- ],
- "prompt_number": 78
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 6.14 Page no 192"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "l=3\n",
- "A=0.02*10**-6 #m**2\n",
- "R=2 #ohm\n",
- "\n",
- "#Calculation\n",
- "a=l/(R*A)\n",
- "\n",
- "#Result\n",
- "print\"Electrical conductivity is\", a*10**-7,\"*10**7 S/m\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Electrical conductivity is 7.5 *10**7 S/m\n"
- ]
- }
- ],
- "prompt_number": 82
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 6.15 Page no 193"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "V=1.8 #V\n",
- "I=1.2 #A\n",
- "l=3 #m\n",
- "A=5.4*10**-6 #m**2\n",
- "\n",
- "#Calculation\n",
- "R=V/I\n",
- "G=1/R\n",
- "a=l/(R*A)\n",
- "\n",
- "#Result\n",
- "print\"Conductance is\",round(G,2),\"S\"\n",
- "print\"Conductivity is\", round(a*10**-5,1),\"*10**5 S/m\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Conductance is 0.67 S\n",
- "Conductivity is 3.7 *10**5 S/m\n"
- ]
- }
- ],
- "prompt_number": 92
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 6.16 Page no 193"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "R1=2.1 #ohm\n",
- "a1=300.0 #K\n",
- "a2=373.0\n",
- "R2=2.7 #ohm\n",
- "\n",
- "#Calculation\n",
- "A=(R2-R1)/(R1*(a2-a1))\n",
- "\n",
- "#Result\n",
- "print\"Temperature coefficient is\", round(A*10**3,3),\"*10**-3 /K\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Temperature coefficient is 3.914 *10**-3 /K\n"
- ]
- }
- ],
- "prompt_number": 99
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 6.17 Page no 193"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "a1=50 #degree\n",
- "R1=6.0 #ohm\n",
- "a2=100 #degree\n",
- "R=7 #ohm\n",
- "\n",
- "#Calculation\n",
- "a=(R-R1)/(R1*(a2-a1))\n",
- "R0=R1/(1+(a*a1))\n",
- "\n",
- "#Result\n",
- "print\"Mean temperature coefficient is\", round(a,4),\"degree/C\"\n",
- "print\"Resistance of the conductor is\",round(R0,3),\"W\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Mean temperature coefficient is 0.0033 degree/C\n",
- "Resistance of the conductor is 5.143 W\n"
- ]
- }
- ],
- "prompt_number": 110
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 6.18 Page no 193"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "V=230 #v\n",
- "a=27 #degree C\n",
- "I1=3.2 #A\n",
- "I2=2.8 #A\n",
- "a1=1.70*10**-4 #degree/C\n",
- "\n",
- "#Calculation\n",
- "R1=V/I1\n",
- "R2=V/I2\n",
- "a2=a+((R2-R1)/(R1*a1))\n",
- "\n",
- "#Result\n",
- "print\"Steady temperature is\", round(a2,2),\"degree C\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Steady temperature is 867.34 degree C\n"
- ]
- }
- ],
- "prompt_number": 117
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 6.21 Page no 193"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "R1=1 #ohm\n",
- "R2=2 \n",
- "R3=3\n",
- "E=12 #V\n",
- "\n",
- "#Calculation\n",
- "Rs=R1+R2+R3\n",
- "I=E/Rs\n",
- "R11=I*R1\n",
- "R12=I*R2\n",
- "R13=I*R3\n",
- "\n",
- "#Result\n",
- "print\"(a) Total resistance of the combination is\",Rs,\"ohm\"\n",
- "print\"(b) Potential drop across R1 is\",R11,\"V\"\n",
- "print\" Potential drop across R2 is\",R12,\"V\"\n",
- "print\" Potential drop across R3 is\",R13,\"V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a) Total resistance of the combination is 6 ohm\n",
- "(b) Potential drop across R1 is 2 V\n",
- " Potential drop across R2 is 4 V\n",
- " Potential drop across R3 is 6 V\n"
- ]
- }
- ],
- "prompt_number": 125
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 6.22 Page no 193"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "R1=2.0 #ohm\n",
- "R2=4.0 #ohm\n",
- "R3=5.0 \n",
- "E=20 #V\n",
- "\n",
- "#calculation\n",
- "Rp=1/(1/R1+1/R2+1/R3)\n",
- "I=E/Rp\n",
- "I1=E/R1\n",
- "I2=E/R2\n",
- "I3=E/R3\n",
- "\n",
- "print\"(a) Toatl resisatnce is\", round(Rp,2),\"ohm\"\n",
- "print\"(b) Current through R1 is\",I1,\"A\"\n",
- "print\"Current through R2 is\",I2,\"A\"\n",
- "print\"Current through R3 is\",I3,\"A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a) Toatl resisatnce is 1.05 ohm\n",
- "(b) Current through R1 is 10.0 A\n",
- "Current through R2 is 5.0 A\n",
- "Current through R3 is 4.0 A\n"
- ]
- }
- ],
- "prompt_number": 134
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 6.24 Page no 194"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "R=6\n",
- "R1=2.0\n",
- "\n",
- "#calculation\n",
- "n=R/R1\n",
- "\n",
- "#Result\n",
- "print\"Number of resistance is\",n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Number of resistance is 3.0\n"
- ]
- }
- ],
- "prompt_number": 135
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 6.25 Page no 194"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "R1=9.0 #ohm\n",
- "R2=5.0\n",
- "R3=3.0\n",
- "R4=7.0\n",
- "\n",
- "#Calculation\n",
- "R11=R1+R2\n",
- "R22=R3+R4\n",
- "R=1/(1/R11+1/R22+1/R22)\n",
- "\n",
- "#Result\n",
- "print\"Equivalent resistance is\",round(R,2),\"ohm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Equivalent resistance is 3.68 ohm\n"
- ]
- }
- ],
- "prompt_number": 141
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 6.26 Page no 194"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "R1=3 #ohm\n",
- "R2=3\n",
- "\n",
- "#Calculation\n",
- "R11=R1+R2\n",
- "R22=(R11*R11)/(R11+R11)\n",
- "R=R11*R1/(R11+R1)\n",
- "\n",
- "#Result\n",
- "print\"Effective resistance is\", R,\"ohm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Effective resistance is 2 ohm\n"
- ]
- }
- ],
- "prompt_number": 148
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 6.27 Page no 195"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "R1=30.0 #ohm\n",
- "R2=30.0\n",
- "E=2 #V\n",
- "\n",
- "#Calculation\n",
- "R11=R1+R2\n",
- "R=R11*R1/(R11+R1)\n",
- "I=E/R\n",
- "\n",
- "#Result\n",
- "print\"Current through the cell is\",I,\"A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Current through the cell is 0.1 A\n"
- ]
- }
- ],
- "prompt_number": 151
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 6.28 Page no 195"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "Rl=500.0 #ohm\n",
- "E=50 #V\n",
- "Rac=2000.0 #ohm\n",
- "Rab=500.0 \n",
- "\n",
- "#Calculation\n",
- "Rbc=Rac-Rab\n",
- "R11=1/(1/Rbc+1/Rl)\n",
- "I=E/(Rab+R11)\n",
- "V=E-(I*Rab)\n",
- "I1=E/Rac\n",
- "Rbc1=40/I1\n",
- "\n",
- "#Result\n",
- "print\"(a) Potential difference across the load is\", round(V,2),\"V\"\n",
- "print\"(b) Resistance is\",Rbc1,\"ohm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a) Potential difference across the load is 21.43 V\n",
- "(b) Resistance is 1600.0 ohm\n"
- ]
- }
- ],
- "prompt_number": 165
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 6.29 Page no 195"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "R1=3 #ohm\n",
- "E=9 #V\n",
- "\n",
- "#Calcuation\n",
- "R=R1+R1+R1\n",
- "I=E/R\n",
- "I1=I/4.0\n",
- "\n",
- "#Result\n",
- "print\"Total resistance is\",R,\"ohm\"\n",
- "print\"Current is\",I1,\"A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Total resistance is 9 ohm\n",
- "Current is 0.25 A\n"
- ]
- }
- ],
- "prompt_number": 169
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 6.30 Page no 196"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "R1=1 #ohm\n",
- "R2=1 \n",
- "r=2/3.0 #ohm\n",
- "E=1 #V\n",
- "\n",
- "#Calculation\n",
- "R11=R1*R2/(R1+R2)\n",
- "R=1/(1/R1+1/R2+1/R1)\n",
- "I=E/(R+r)\n",
- "\n",
- "#Result\n",
- "print\"Current drawn from the cell is\", I,\"A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Current drawn from the cell is 1.5 A\n"
- ]
- }
- ],
- "prompt_number": 174
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 6.31 Page no 196"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "R=10.0 #ohm\n",
- "r=3.33\n",
- "E=15\n",
- "\n",
- "#Calculation\n",
- "Rp=1/(1/R+(1/(R+R)))\n",
- "R1=R+Rp+R\n",
- "I=E/(R1+r)\n",
- "V=I*Rp\n",
- "I1=V/(R+R)\n",
- "I2=V/R\n",
- "\n",
- "#Result\n",
- "print\"Current flow through the main circuit is\", round(I,1),\"A\"\n",
- "print\"Current through QRS is\",round(I1,3),\"A\"\n",
- "print\"Current through QS is\",round(I2,3),\"A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Current flow through the main circuit is 0.5 A\n",
- "Current through QRS is 0.167 A\n",
- "Current through QS is 0.333 A\n"
- ]
- }
- ],
- "prompt_number": 199
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 6.32 Page no 196"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "R1=2 #ohm\n",
- "R2=3 \n",
- "R3=2.8\n",
- "E=6 #V\n",
- "\n",
- "#calculation\n",
- "R=(R1*R2/(R1+R2))+R3\n",
- "I=E/R\n",
- "\n",
- "#Result\n",
- "print\"Current in steady state is\", round(I,1),\"A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Current in steady state is 1.6 A\n"
- ]
- }
- ],
- "prompt_number": 206
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 6.33 Page no 197"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "E=10 #v\n",
- "r=3 #ohm\n",
- "I=0.5 #A\n",
- "\n",
- "#Calculation\n",
- "R=(E/I)-r\n",
- "V=I*R\n",
- "\n",
- "#Result\n",
- "print\"Resistance of the resistor is\", R,\"ohm\"\n",
- "print\"Terminal voltage is\",V,\"V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Resistance of the resistor is 17.0 ohm\n",
- "Terminal voltage is 8.5 V\n"
- ]
- }
- ],
- "prompt_number": 212
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 6.35 Page no 197"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "R1=3.0 #ohm\n",
- "R2=4.0\n",
- "R3=6.0\n",
- "r=2/3.0\n",
- "E=2 #V\n",
- "\n",
- "#Calculation\n",
- "Rp=1/(1/R1+1/R2+1/R3)\n",
- "R=Rp+r\n",
- "I=E/R\n",
- "V=I*Rp\n",
- "I1=V/R1\n",
- "\n",
- "#Result\n",
- "print\"Current drawn from the cell is\", I,\"A\"\n",
- "print\"Current through 3 Ohm resistance is\",round(I1,2),\"A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Current drawn from the cell is 1.0 A\n",
- "Current through 3 Ohm resistance is 0.44 A\n"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 6.36 Page no 197"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "E=8\n",
- "R=3.0\n",
- "r=1.0\n",
- "\n",
- "#Calculation\n",
- "I=E/(R+r)\n",
- "I1=I/2.0\n",
- "\n",
- "#Result\n",
- "print\"Current from the battery is\",I,\"A\"\n",
- "print\"Current through AC and ABC is\",I1,\"A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Current from the battery is 2.0 A\n",
- "Current through AC and ABC is 1.0 A\n"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 6.37 Page no 198"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "E=12 #V\n",
- "R2=8 #ohm\n",
- "r=1 #ohm\n",
- "\n",
- "#Calculation\n",
- "R1=-((E/I)-R2-r)\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 3.0 ohm\n"
- ]
- }
- ],
- "prompt_number": 17
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 6.38 Page no 198"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "E1=20 #V\n",
- "r1=1 #ohm\n",
- "E2=8 #Volt\n",
- "r2=2 #ohm\n",
- "R1=12.0\n",
- "R2=6.0\n",
- "R3=4.0\n",
- "\n",
- "#Calculation\n",
- "E=E1-E2\n",
- "R11=1/(1/R1+1/R2+1/R3)\n",
- "R=R11+r1+r2+5\n",
- "I=E/R\n",
- "I1=I-1\n",
- "V1=E1-(I*r1)\n",
- "V2=E2+(I*r2)\n",
- "\n",
- "#Result\n",
- "print\"(a) Current in the circuit is\", I,\"A\"\n",
- "print\"(b) Current in the resister is\", I1,\"A\"\n",
- "print\"(c) Potential difference across 12 V battery is\", V1,\"V\"\n",
- "print\" Potential difference across 8 V battery is\",V2,\"V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a) Current in the circuit is 1.2 A\n",
- "(b) Current in the resister is 0.2 A\n",
- "(c) Potential difference across 12 V battery is 18.8 V\n",
- " Potential difference across 8 V battery is 10.4 V\n"
- ]
- }
- ],
- "prompt_number": 34
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 6.39 Page no 198"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "E=2 #V\n",
- "r=0.015 #ohm\n",
- "R=8.5\n",
- "\n",
- "#Calculation\n",
- "E1=E*6\n",
- "r1=r*6\n",
- "I=E1/(R+r1)\n",
- "V=I*R\n",
- "\n",
- "#Result\n",
- "print\"Current drawn from the supply is\", round(I,3),\"A\"\n",
- "print\"Terminal voltage is\",round(V,3),\"V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Current drawn from the supply is 1.397 A\n",
- "Terminal voltage is 11.874 V\n"
- ]
- }
- ],
- "prompt_number": 41
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 6.40 Page no 198"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "E=4 #V\n",
- "I=1.5 #A\n",
- "R=6 #ohm\n",
- "\n",
- "#Calculation\n",
- "E1=3*E\n",
- "r=((E1/I)-R)/3.0\n",
- "V=E-(I*r)\n",
- "\n",
- "#Result\n",
- "print\"(a) Internal resistance is\", round(r,2),\"ohm\"\n",
- "print\"(b) Terminal voltage is\",V,\"Volt\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a) Internal resistance is 0.67 ohm\n",
- "(b) Terminal voltage is 3.0 Volt\n"
- ]
- }
- ],
- "prompt_number": 48
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 6.41 Page no 199"
- ]
- },
- {
- "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 resistance is\", r,\"ohm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Internal resistance is 7.5 ohm\n"
- ]
- }
- ],
- "prompt_number": 53
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 6.42 Page no 199"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "E=1\n",
- "a=2\n",
- "\n",
- "#Calculation\n",
- "r=(a*E*(a-E))/(a*E*(a-E))\n",
- "\n",
- "#Result\n",
- "print\"The internal resistance is\",r,\"ohm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The internal resistance is 1 ohm\n"
- ]
- }
- ],
- "prompt_number": 66
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 6.43 Page no 199"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "E=2 #V\n",
- "r=1.5 #ohm\n",
- "R=10 #ohm\n",
- "r1=1.5\n",
- "r2=1.5\n",
- "\n",
- "#Calculation\n",
- "E1=2*E\n",
- "w=r1+r2\n",
- "A=1/(1/w+1/w)\n",
- "B=R+A\n",
- "I=E1/B\n",
- "IR=I*R\n",
- "\n",
- "#Result\n",
- "print\"The potential difference across external resistance is\",round(IR,2),\"V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The potential difference across external resistance is 3.48 V\n"
- ]
- }
- ],
- "prompt_number": 63
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 6.44 Page no 200"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "n=2.7*10**18\n",
- "N=1.0*10**18\n",
- "e=1.6*10**-19\n",
- "V=230 #V\n",
- "\n",
- "#Calculation\n",
- "I=(n+N)*(e)\n",
- "R=V/I\n",
- "\n",
- "#Result\n",
- "print\"The effective resistance of the tube is\",round(R,1),\"ohm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The effective resistance of the tube is 388.5 ohm\n"
- ]
- }
- ],
- "prompt_number": 73
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 6.47 Page no 200"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "X=1\n",
- "a=3\n",
- "E1=12\n",
- "r=0.5\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "X1=X+math.sqrt(a)\n",
- "I=E1/(X1+r)\n",
- "\n",
- "#Result\n",
- "print\"Current drawn is\", round(I,3),\"A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Current drawn is 3.713 A\n"
- ]
- }
- ],
- "prompt_number": 99
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 6.48 Page no 200"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "R1=6\n",
- "R2=3.0\n",
- "R3=5\n",
- "E=24 #V\n",
- "I=3 #A\n",
- "\n",
- "#Calculation\n",
- "Rp=(R1*R2)/(R1+R2)\n",
- "R=R3+Rp+r\n",
- "r=R-(R3+Rp)\n",
- "V=I*Rp\n",
- "I1=V/R1\n",
- "I2=V/R2\n",
- "R11=R2+R3+r\n",
- "I11=E/R11\n",
- "\n",
- "#Result\n",
- "print\"(a) Internal resistance is\", r,\"ohm\"\n",
- "print\" Current I1 is\",I1,\"A and I2 is\",I2,\"A\"\n",
- "print\"(b) Current in the circuit is\",round(I11,2),\"A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a) Internal resistance is 1.0 ohm\n",
- " Current I1 is 1.0 A and I2 is 2.0 A\n",
- "(b) Current in the circuit is 2.67 A\n"
- ]
- }
- ],
- "prompt_number": 91
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/modern_physics_by_Satish_K._Gupta/chap7.ipynb b/modern_physics_by_Satish_K._Gupta/chap7.ipynb deleted file mode 100755 index 0476eb75..00000000 --- a/modern_physics_by_Satish_K._Gupta/chap7.ipynb +++ /dev/null @@ -1,832 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:3ec304c8c9eb23fdd2ae23967a8effd62df3acdefa633ca8bf2a83485250a21b"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 7 Electrical Measurements"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 7.1 Page no 229"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "R1=2\n",
- "R2=3\n",
- "R3=5\n",
- "E1=6\n",
- "E2=4\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "A=np.array([[7,5],[5,8]])\n",
- "b=np.array([6,4])\n",
- "z=np.linalg.solve(A,b)\n",
- "Z=z[0]\n",
- "Z1=z[1]\n",
- "A1=Z+Z1\n",
- "\n",
- "#Result\n",
- "print\"Current through R1 is\", round(Z,3),\"A\"\n",
- "print\"Current through R2 is\", round(Z1,3),\"A\"\n",
- "print\"Current through R3 is\",round(A1,2),\"A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Current through R1 is 0.903 A\n",
- "Current through R2 is -0.065 A\n",
- "Current through R3 is 0.84 A\n"
- ]
- }
- ],
- "prompt_number": 33
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 7.2 Page no 229"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "E1=1.5 #V\n",
- "E2=2.0 #V\n",
- "r1=1 #ohm\n",
- "r2=2 #ohm\n",
- "R=5 #ohm\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "A=np.array([[6,5],[5,7]])\n",
- "B=np.array([1.5,2])\n",
- "Z=np.linalg.solve(A,B)\n",
- "Z1=Z[0]\n",
- "Z2=Z[1]\n",
- "A1=(Z1+Z2)*R\n",
- "\n",
- "#Result\n",
- "print\"Current through R1 is\", round(Z1,4),\"A\"\n",
- "print\"Current through R2 is\",round(Z2,4),\"A\"\n",
- "print\"Current through R3 is\",round(A1,4),\"A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Current through R1 is 0.0294 A\n",
- "Current through R2 is 0.2647 A\n",
- "Current through R3 is 1.4706 A\n"
- ]
- }
- ],
- "prompt_number": 42
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 7.3 Page no 230"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "E1=2 #V\n",
- "E2=1 #V\n",
- "E3=4 #V\n",
- "r1=4 #ohm\n",
- "r2=3 #ohm\n",
- "r3=2 #ohm\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "A=np.array([[1,1,1],[4,-3,0],[0,3,-2]])\n",
- "A1=np.array([0,1,-3])\n",
- "Z=np.linalg.solve(A,A1)\n",
- "Z1=Z[0]\n",
- "Z2=Z[1]\n",
- "Z3=Z[2]\n",
- "\n",
- "#Result\n",
- "print\"Current through R1 is\", round(Z1,2),\"A\"\n",
- "print\"Current through R2 is\",round(Z2,2),\"A\"\n",
- "print\"Current through R3 is\",round(Z3,2),\"A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Current through R1 is -0.15 A\n",
- "Current through R2 is -0.54 A\n",
- "Current through R3 is 0.69 A\n"
- ]
- }
- ],
- "prompt_number": 26
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 7.4 Page no 230"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "R1=5 #ohm\n",
- "R3=2 \n",
- "R=10\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "A=np.array([[9,5],[10,19]])\n",
- "A1=np.array([1,1])\n",
- "Z=np.linalg.solve(A,A1)\n",
- "Z1=Z[0]\n",
- "Z2=Z[1]\n",
- "I=(Z1+Z2)*R\n",
- "\n",
- "#Result\n",
- "print\"Potential difference across the resistor is\", round(I,4),\"V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Potential difference across the resistor is 1.0744 V\n"
- ]
- }
- ],
- "prompt_number": 25
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 7.5 Page no 230"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "A=10 #ohm\n",
- "B=5 #ohm\n",
- "S=2\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "A=np.array([[25,-10],[5,5]])\n",
- "B=np.array([9,7]) \n",
- "Z=np.linalg.solve(A,B)\n",
- "Z1=Z[1]*10\n",
- "\n",
- "#Result\n",
- "print\"Equivalent resistance is\", round(Z1,0),\"ohm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Equivalent resistance is 7.0 ohm\n"
- ]
- }
- ],
- "prompt_number": 51
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 7.6 Page no 231 "
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "P=15 #ohm\n",
- "Q=12.0\n",
- "R=10.0\n",
- "S=4\n",
- "\n",
- "#Calculation\n",
- "R1=S*(P/Q)\n",
- "X=1/(1/R1-1/R)\n",
- "\n",
- "#Result\n",
- "print\"The resistance to be connectedin parallel is\",X,\"ohm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The resistance to be connectedin parallel is 10.0 ohm\n"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 7.7 Page no 231"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "P=4.0\n",
- "Q=4.0\n",
- "R=4.0\n",
- "X=4.0\n",
- "\n",
- "#Calculation\n",
- "R1=P+Q\n",
- "R2=R+X\n",
- "R3=1/(1/R1+1/R2)\n",
- "I=P/R3\n",
- "I1=I/2\n",
- "\n",
- "#Result\n",
- "print\"(i) Equivalent resistance is\",R3,\"ohm\"\n",
- "print\"(ii) The magnitudes of current is\",I1,\"A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(i) Equivalent resistance is 4.0 ohm\n",
- "(ii) The magnitudes of current is 0.5 A\n"
- ]
- }
- ],
- "prompt_number": 41
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 7.8 Page no 232"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "a=4\n",
- "b=3\n",
- "c=2.0\n",
- "E=6 #v\n",
- "\n",
- "#Calculation\n",
- "X=(a*b)/c\n",
- "R1=a+c\n",
- "R2=b+R1\n",
- "R=((R1*R2)/(R1+R2))+2.4\n",
- "I=E/R\n",
- "\n",
- "#Result\n",
- "print\"Current drawn by the circuit is\", I,\"A\"\n",
- "print\"Unknown resistance is\",R,\"ohm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Current drawn by the circuit is 1.0 A\n",
- "Unknown resistance is 6.0 ohm\n"
- ]
- }
- ],
- "prompt_number": 59
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 7.9 Page no 232"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "a=1.5\n",
- "b=5\n",
- "c=6\n",
- "\n",
- "#Calculation\n",
- "I1=a/(b+c)\n",
- "I2=I1/10.0\n",
- "I=I1+I2\n",
- "\n",
- "#Result\n",
- "print\"Current in the arms is\", round(I1,4),\"A and\",round(I2,4),\"A\"\n",
- "print\"Current in the cell is\",I,\"A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Current in the arms is 0.1364 A and 0.0136 A\n",
- "Current in the cell is 0.15 A\n"
- ]
- }
- ],
- "prompt_number": 68
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 7.10 Page no 233"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "X=10\n",
- "P=3\n",
- "Q=2.0\n",
- "\n",
- "#Calculation\n",
- "R=X*P/Q\n",
- "L=1/R\n",
- "\n",
- "#Result\n",
- "print\"Length is\",round(L,3),\"m\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Length is 0.067 m\n"
- ]
- }
- ],
- "prompt_number": 71
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 7.11 Page no 233"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "r=6 #ohm\n",
- "l=1/3.0\n",
- "r1=2/3.0\n",
- "\n",
- "#Calculation\n",
- "R1=r/((2.0*(r1/l))-1)\n",
- "R2=2*R1\n",
- "\n",
- "#Result\n",
- "print\"Resistance of two wires is\", R1,\"ohm and\",R2,\"ohm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Resistance of two wires is 2.0 ohm and 4.0 ohm\n"
- ]
- }
- ],
- "prompt_number": 76
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 7.12 Page no 233"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "P=1.2 #ohm\n",
- "Q=0.8\n",
- "R=2\n",
- "E=4 #V\n",
- "\n",
- "#Calculation\n",
- "X=(R*P)/Q\n",
- "R1=X+R\n",
- "R2=P+Q\n",
- "R11=R1*R2/(R1+R2)\n",
- "I=E/R11\n",
- "\n",
- "#Result\n",
- "print\"Unknown resistance is\", X,\"ohm\"\n",
- "print\"Current drawn is\",I,\"A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Unknown resistance is 3.0 ohm\n",
- "Current drawn is 2.8 A\n"
- ]
- }
- ],
- "prompt_number": 83
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 7.13 Page no 233"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "R=20.0 #ohm\n",
- "V=10**-3 #V/m\n",
- "l=10**4 #mm\n",
- "V1=10**-2 #V\n",
- "\n",
- "#Calculation\n",
- "I=V1/R\n",
- "R1=(2/I)-R\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 3980.0 ohm\n"
- ]
- }
- ],
- "prompt_number": 87
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 7.14 Page no 233"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "l1=65 #cm\n",
- "l2=60.0\n",
- "a=0.1\n",
- "\n",
- "#Calculation\n",
- "E1=((a*l1)/l2)/((l1/l2)-1)\n",
- "E2=E1-a\n",
- "\n",
- "#Result\n",
- "print\"E.m.f of cell is\", E1,\"V and\",E2,\"V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "E.m.f of cell is 1.3 V and 1.2 V\n"
- ]
- }
- ],
- "prompt_number": 93
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 7.15 Page no 234"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "S=9.5 #ohm\n",
- "I1=76.3 #cm\n",
- "l2=64.8\n",
- "\n",
- "#Calculation\n",
- "r=((I1/I2)-1)*S\n",
- "\n",
- "#Result\n",
- "print\"Internal resistance is\",round(r*10**-3,1),\"ohm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Internal resistance is 53.1 ohm\n"
- ]
- }
- ],
- "prompt_number": 138
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 7.16 Page no 234"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "E=10*10**-3 \n",
- "r=10\n",
- "l=40 #cm\n",
- "\n",
- "#Calculation\n",
- "R=10*l/100.0\n",
- "R1=(R*2/E)-10\n",
- "\n",
- "#Result\n",
- "print\"External resistance is\", R1,\"ohm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "External resistance is 790.0 ohm\n"
- ]
- }
- ],
- "prompt_number": 111
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 7.17 Page no 234"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "R1=10*10**3\n",
- "E=12\n",
- "\n",
- "#Calculation\n",
- "R=R1/((E/2.0)-1)\n",
- "\n",
- "#Result\n",
- "print\"(a) Resistance is\",R*10**-3,\"K ohm\"\n",
- "print\"(b) When current flows through a resistor,its temperature increases due toheat produced across it.\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a) Resistance is 2.0 K ohm\n",
- "(b) When current flows through a resistor,its temperature increases due toheat produced across it.\n"
- ]
- }
- ],
- "prompt_number": 116
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 7.18 Page no 235"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "a=3.0\n",
- "b=6.0\n",
- "\n",
- "#Caculation\n",
- "E=(1/a)+(1/b)+1+(1/a)\n",
- "\n",
- "#Result\n",
- "print\"Equivalent resistance is\",round(E,2),\"r\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Equivalent resistance is 1.83 r\n"
- ]
- }
- ],
- "prompt_number": 120
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 7.22 Page no 237"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "a=5\n",
- "b=10\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "A=np.array([[7,1],[1,5]])\n",
- "A1=np.array([2,2])\n",
- "B=np.linalg.solve(A,A1)\n",
- "B1=B[0]\n",
- "B2=B[1]\n",
- "E=(a*B2)+(B1*b)\n",
- "\n",
- "#Result\n",
- "print\"Equivalent resistance is\", round(E,2),\"ohm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Equivalent resistance is 4.12 ohm\n"
- ]
- }
- ],
- "prompt_number": 125
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 7.23 Page no 238"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "R=12.5\n",
- "l=39.5\n",
- "\n",
- "#Calculation\n",
- "X=(R*(100-l))/l\n",
- "L=X/(X+R)\n",
- "\n",
- "#Result\n",
- "print\"(a) Resistance is\",round(X,2),\"W\"\n",
- "print\"(b) Balance point of bridge is\", L*10**2,\"cm\"\n",
- "print\"(c) The galvanometer will not show any current.\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a) Resistance is 19.15 W\n",
- "(b) Balance point of bridge is 60.5 cm\n",
- "(c) The galvanometer will not show any current.\n"
- ]
- }
- ],
- "prompt_number": 135
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/modern_physics_by_Satish_K._Gupta/chap8.ipynb b/modern_physics_by_Satish_K._Gupta/chap8.ipynb deleted file mode 100755 index 1b06633c..00000000 --- a/modern_physics_by_Satish_K._Gupta/chap8.ipynb +++ /dev/null @@ -1,994 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:a40c0c75d3ca70fc3ecf4f82797b161ff3f00d80a8c81333303a3e01cdc2b09b"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 8 Thermal effects of current"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 8.1 Page no 258"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "P=100.0 #W\n",
- "V=220.0 #V\n",
- "\n",
- "#Calculation\n",
- "I=P/V\n",
- "R=V**2/P\n",
- "\n",
- "#Result\n",
- "print\"Resistance is\",R,\"ohm\"\n",
- "print\"Current capacity is\",round(I,3),\"A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Resistance is 484.0 ohm\n",
- "Current capacity is 0.455 A\n"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 8.2 Page no 258"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "e=1.6*10**-19 #C\n",
- "P=100 #W\n",
- "V=200.0 #V\n",
- "t=1\n",
- "\n",
- "#Calculation\n",
- "I=P/V\n",
- "q=I*t\n",
- "N=q/e\n",
- "\n",
- "#Result\n",
- "print\"Number of electrons is\", N"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Number of electrons is 3.125e+18\n"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 8.3 Page no 258"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "P1=60 #W\n",
- "P2=100.0 \n",
- "\n",
- "#Calculation\n",
- "I=P1/P2\n",
- "\n",
- "#Result\n",
- "print\"Hence 100 W bulb draws more current\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Hence 100 W bulb draws more current\n"
- ]
- }
- ],
- "prompt_number": 10
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 8.4 Page no 259"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "P1=60.0 #W\n",
- "P2=100\n",
- "\n",
- "#Calculation\n",
- "R=P2/P1\n",
- "\n",
- "#Result\n",
- "print\"(a) 60 W bulb will dissipate more power.\"\n",
- "print\"(b) Since 60 W bulb is dissipating more power, it glows brighter.\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a) 60 W bulb will dissipate more power.\n",
- "(b) Since 60 W bulb is dissipating more power, it glows brighter.\n"
- ]
- }
- ],
- "prompt_number": 12
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 8.5 Page no 259"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "P1=500.0 #W\n",
- "V1=220 #V\n",
- "P2=100.0\n",
- "V2=220\n",
- "\n",
- "#Calculation\n",
- "R1=V1**2/P1\n",
- "R2=V2**2/P2\n",
- "I=V2/(R1+R2)\n",
- "P11=I**2*R1\n",
- "P22=I**2*R2\n",
- "\n",
- "#Result\n",
- "print\"(i) Power consumed by heater is\",round(P11,2),\"W\"\n",
- "print\"(ii) Power consumed by electric bulb is\",round(P22,2),\"W\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(i) Power consumed by heater is 13.89 W\n",
- "(ii) Power consumed by electric bulb is 69.44 W\n"
- ]
- }
- ],
- "prompt_number": 18
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 8.6 Page no 259"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "P=100.0 #W\n",
- "V=220 #V\n",
- "V1=110\n",
- "\n",
- "#Calculation\n",
- "R=V**2/P\n",
- "P1=V1**2/R\n",
- "\n",
- "#Result\n",
- "print\"Power consumed by the bulb is\",P1,\"W\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Power consumed by the bulb is 25.0 W\n"
- ]
- }
- ],
- "prompt_number": 23
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 8.7 Page no 259"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "P=550 #W\n",
- "V=220 #V\n",
- "V1=222.2 #V\n",
- "V2=217.8 #V\n",
- "\n",
- "#Calculation\n",
- "R=V**2/P\n",
- "P=V1**2/R\n",
- "P1=V2**2/R\n",
- "\n",
- "#Result\n",
- "print\"minimum power is\",round(P,0),\"W\"\n",
- "print\"maximum power is\",round(P1,0),\"W\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "minimum power is 561.0 W\n",
- "maximum power is 539.0 W\n"
- ]
- }
- ],
- "prompt_number": 30
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 8.8 Page no 259"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "V=220\n",
- "P=60.0 #W\n",
- "P1=85\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "R=V**2/P\n",
- "V1=math.sqrt(P1*R)\n",
- "\n",
- "#Result\n",
- "print\"Voltage fluctuation is\",round(V1,1),\"V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Voltage fluctuation is 261.9 V\n"
- ]
- }
- ],
- "prompt_number": 34
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 8.9 Page no 259"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "P=50*10**3 #W\n",
- "V=5000.0\n",
- "R=20\n",
- "\n",
- "#Calculation\n",
- "I=P/V\n",
- "P1=I**2*R\n",
- "P2=P-P1\n",
- "\n",
- "#Result \n",
- "print\"Power received by the factory is\",P2*10**-3,\"Kw\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Power received by the factory is 48.0 Kw\n"
- ]
- }
- ],
- "prompt_number": 42
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 8.10 Page no 260"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "R2=2 #ohm\n",
- "R3=2 #ohm\n",
- "I=3 #A\n",
- "\n",
- "#Calculation\n",
- "P=I**2*R2\n",
- "R1=(R2*R3)/(R2+R3)\n",
- "P1=I**2*R1\n",
- "D=P+P1\n",
- "\n",
- "#Result\n",
- "print\"The maximum power is\",D,\"W\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The maximum power is 27 W\n"
- ]
- }
- ],
- "prompt_number": 50
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 8.11 Page no 260"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "R=0.5 #ohm\n",
- "I=100 #A\n",
- "T=24*60*60\n",
- "\n",
- "#Calculation\n",
- "Q=I**2*R*T\n",
- "\n",
- "#Result\n",
- "print\"Energy is being lost per day in the form of heat is\",Q*10**-8,\"*10**8 J\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Energy is being lost per day in the form of heat is 4.32 *10**8 J\n"
- ]
- }
- ],
- "prompt_number": 59
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 8.12 Page no 260"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "R=50.0 #ohm\n",
- "V=10 #V\n",
- "T=3600 #s\n",
- "\n",
- "#Calculation\n",
- "I=V/R\n",
- "W=I**2*R*T\n",
- "\n",
- "#Result \n",
- "print\"The source of this energy is\",W,\"J\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The source of this energy is 7200.0 J\n"
- ]
- }
- ],
- "prompt_number": 64
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 8.13 Page no 260"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "P=100 #W\n",
- "V=200 #V\n",
- "T=60\n",
- "I=10\n",
- "\n",
- "#Calculation\n",
- "R=V**2/P\n",
- "W=(P**2/R)*I*T\n",
- "\n",
- "#Result \n",
- "print\"The heat & light produced by the bulb is\",W,\"J\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The heat & light produced by the bulb is 15000 J\n"
- ]
- }
- ],
- "prompt_number": 70
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 8.14 Page no 260"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "R1=3 #ohm\n",
- "R2=6\n",
- "R3=2\n",
- "V=12 #V\n",
- "T=300 #s\n",
- "\n",
- "#Calculation\n",
- "R12=R1*R2/(R1+R2)\n",
- "R=R12+R3\n",
- "I=V/R\n",
- "I1=I*R2/(R1+R2)\n",
- "I2=I*R1/(R1+R2)\n",
- "Q1=I1**2*R1*T\n",
- "Q2=I2**2*R2*T\n",
- "Q3=I**2*R3*T\n",
- "\n",
- "#Result \n",
- "print\"Heat produced in resistance R1 is\",Q1,\"J\"\n",
- "print\"Heat produced in resistance R2 is\",Q2,\"J\"\n",
- "print\"Heat produced in resistance R3 is\",Q3,\"J\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Heat produced in resistance R1 is 3600 J\n",
- "Heat produced in resistance R2 is 1800 J\n",
- "Heat produced in resistance R3 is 5400 J\n"
- ]
- }
- ],
- "prompt_number": 82
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 8.15 Page no 260"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "t1=300 #S\n",
- "t2=600 #S\n",
- "\n",
- "#Calculation\n",
- "t=(t1+t2)/60.0\n",
- "\n",
- "#Result\n",
- "print\"Time required to boil water is\",t,\"minute\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Time required to boil water is 15.0 minute\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 8.16 Page no 261"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "R=100 #ohm\n",
- "I=0.5 #A\n",
- "W=10*10**-3 #Kg\n",
- "M=250*10**-3 #Kg\n",
- "T=1800 #s\n",
- "C=4.2*10**3\n",
- "\n",
- "#Calculation\n",
- "Q=I**2*R*T\n",
- "V=Q/((W+M)*C)\n",
- "\n",
- "#Result\n",
- "print\"The rise of the temperature is\",round(V,1),\"C\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The rise of the temperature is 41.2 C\n"
- ]
- }
- ],
- "prompt_number": 14
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 8.17 Page no 261"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "V=200\n",
- "R=100\n",
- "R1=121.0\n",
- "A=1000 #W\n",
- "B=400 #W\n",
- "C=400 #W\n",
- "T=116 #h\n",
- "G=2.50\n",
- "\n",
- "#Calculation\n",
- "D=V**2/R\n",
- "D1=V**2/R1\n",
- "E=A+B+C+D1\n",
- "F=E*T\n",
- "H=F*G\n",
- "\n",
- "#Result \n",
- "print\"The bill for february is\",round(H*10**-3,2)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The bill for february is 617.87\n"
- ]
- }
- ],
- "prompt_number": 26
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 8.18 Page no 261"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "a=0.12 #KWh\n",
- "b=1.44 #KWh\n",
- "c=0.48 #KWh\n",
- "d=0.24 \n",
- "e=0.30\n",
- "i=2.70\n",
- "g=30\n",
- "\n",
- "#Calculation\n",
- "f=a+b+c+d+e\n",
- "h=2*g*f\n",
- "k=i*h\n",
- "\n",
- "#Result\n",
- "print\"Electric bill for 2 month is\", k"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Electric bill for 2 month is 417.96\n"
- ]
- }
- ],
- "prompt_number": 103
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 8.19 Page no 261"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "P=100*10**6 #W\n",
- "V1=20000\n",
- "V2=200 #V\n",
- "\n",
- "#Calculation\n",
- "I1=P/V1\n",
- "Q1=I1**2\n",
- "I2=P/V2\n",
- "Q2=I2**2\n",
- "\n",
- "#Result\n",
- "print\"Transmission done at V1=20,000 is\", Q1*10**-5,\"10**5\"\n",
- "print\"Transmission done at V2=200 V is\", Q2*10**-10,\"10**10\"\n",
- "print\"Less power wastage when transmission is done at Q1\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Transmission done at V1=20,000 is 250.0 10**5\n",
- "Transmission done at V2=200 V is 25.0 10**10\n",
- "Less power wastage when transmission is done at Q1\n"
- ]
- }
- ],
- "prompt_number": 105
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 8.20 Page no 262"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "p1=4.9*10**-7 #ohm m\n",
- "l1=8.456 #m\n",
- "b1=1.0 #mm\n",
- "t1=0.03*10**-6 #m\n",
- "l2=4.235\n",
- "A2=0.12*10**-6\n",
- "p2=1.1*10**-6\n",
- "\n",
- "#Calculation\n",
- "A1=b1*t1\n",
- "R1=p1*(l1/A1)\n",
- "R2=p2*(l2/A2)\n",
- "\n",
- "#Result\n",
- "print\"Rate for heat production for constantan ribbon is\", round(R1,3),\"W\"\n",
- "print\"Rate for heat production for nichrome ribbon is\",round(R2,3),\"W\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Rate for heat production for constantan ribbon is 138.115 W\n",
- "Rate for heat production for nichrome ribbon is 38.821 W\n"
- ]
- }
- ],
- "prompt_number": 115
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 8.21 Page no 262"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "V=50 #V\n",
- "I=12 #A\n",
- "d=70\n",
- "\n",
- "#Calculation\n",
- "g=V*I\n",
- "t=g*(d/100.0)\n",
- "v=t/I**2\n",
- "\n",
- "#Result \n",
- "print\"The resistance of the motor is\",round(v,2),\"ohm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The resistance of the motor is 2.92 ohm\n"
- ]
- }
- ],
- "prompt_number": 76
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 8.22 Page no 262"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "a=100\n",
- "b=6\n",
- "c=8.0\n",
- "d=2.0\n",
- "f=0.50\n",
- "t=15*60\n",
- "\n",
- "#Calculation\n",
- "e=a-b*d\n",
- "x=b*f\n",
- "r=(e/c)-x\n",
- "r1=a*r\n",
- "r2=r**2*(r+b*f)\n",
- "r3=r1-r2\n",
- "M=r3*t\n",
- "\n",
- "#Result\n",
- "print\"Resistance in the charging circuit is\",r,\"ohm\"\n",
- "print\"The power supplied by the d.c source is\",r1,\"W\"\n",
- "print\"The power dissipated as heat is\",r2,\"W\"\n",
- "print\"The power stored in the battery is\",r3,\"W\"\n",
- "print\"Energy stored in the battery in 15 minute is\",M,\"J\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Resistance in the charging circuit is 8.0 ohm\n",
- "The power supplied by the d.c source is 800.0 W\n",
- "The power dissipated as heat is 704.0 W\n",
- "The power stored in the battery is 96.0 W\n",
- "Energy stored in the battery in 15 minute is 86400.0 J\n"
- ]
- }
- ],
- "prompt_number": 96
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 8.23 Page no 262"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "V=64 #V\n",
- "e=2.0 #V\n",
- "n=8\n",
- "r=1/8.0\n",
- "R=7 #ohm\n",
- "I=3.5 #A\n",
- "b=3600\n",
- "\n",
- "#Calculation\n",
- "Te=e*n\n",
- "R1=r*n\n",
- "E=V*I\n",
- "E1=Te*I\n",
- "E11=I**2*(R+R1)\n",
- "Em=E-(E1+E11)\n",
- "Em1=Em*b\n",
- "Ec=E1*b\n",
- "\n",
- "#Result\n",
- "print\"(a) Mechanical energy yielded by motor is\", Em1,\"J\"\n",
- "print\"(b) Chemical energy stored in the battery is\",Ec,\"J\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a) Mechanical energy yielded by motor is 252000.0 J\n",
- "(b) Chemical energy stored in the battery is 201600.0 J\n"
- ]
- }
- ],
- "prompt_number": 130
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 8.25 Page no 263"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "E=24 #V\n",
- "r=4 #ohm\n",
- "a=1\n",
- "\n",
- "#Calculation\n",
- "R=r/(2.0-a)\n",
- "I=E/(R+r)\n",
- "\n",
- "#Result\n",
- "print\"Current drawn from the battery is\", I,\"A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Current drawn from the battery is 3.0 A\n"
- ]
- }
- ],
- "prompt_number": 134
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/modern_physics_by_Satish_K._Gupta/chap9.ipynb b/modern_physics_by_Satish_K._Gupta/chap9.ipynb deleted file mode 100755 index ecaec460..00000000 --- a/modern_physics_by_Satish_K._Gupta/chap9.ipynb +++ /dev/null @@ -1,627 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:3404f51fc9045816d89b5507acdfe9797796d47a0805fc064b3ee38cacd900d7"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 9 Chemical effects of current"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 9.1 Page no 285"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "m=1*10**-3 #kg\n",
- "I=2 #A\n",
- "z=3.3*10**-7 #kg/C\n",
- "\n",
- "#Calculation\n",
- "t=m/(z*I)\n",
- "\n",
- "#Result\n",
- "print\"Time required is\", round(t,1),\"s\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Time required is 1515.2 s\n"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 9.2 Page no 285"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "m=0.15*10**-3 #Kg\n",
- "z=3.3*10**-7 #Kg/C\n",
- "t=900 #S\n",
- "I1=0.6 #A\n",
- "\n",
- "#Calculation\n",
- "I=m/(z*t)\n",
- "I2=I-I1\n",
- "\n",
- "#Result\n",
- "print\"Correction required for the ammeter reading is\", round(I2,1),\"A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Correction required for the ammeter reading is -0.1 A\n"
- ]
- }
- ],
- "prompt_number": 10
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 9.3 Page no 285"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "t=0.002 #m\n",
- "A=72 #cm**2\n",
- "d=8.9 #g/cm**3\n",
- "z=33*10**-5 #g/C\n",
- "I=5 #A\n",
- "\n",
- "#Calculation\n",
- "V=t*A\n",
- "m=V*d\n",
- "t1=m/(z*I)\n",
- "\n",
- "#Result\n",
- "print\"Time required is\", round(t1,0),\"S\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Time required is 777.0 S\n"
- ]
- }
- ],
- "prompt_number": 17
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 9.4 Page no 286"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "m1=2 #g\n",
- "m2=1 #g\n",
- "t=1800 #s\n",
- "z1=1118*10**-6 \n",
- "z2=3294*10**-7 \n",
- "a=6\n",
- "\n",
- "#Calculation\n",
- "l1=m1/(z1*t)\n",
- "l2=m2/(z2*t)\n",
- "l=l1+l2\n",
- "p=l*a\n",
- "\n",
- "#Result\n",
- "print\"Power of the current is\",round(p,3),\"W\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Power of the current is 16.082 W\n"
- ]
- }
- ],
- "prompt_number": 25
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 9.5 Page no 286"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "m1=0.403*10**-3 #Kg\n",
- "z1=1.12*10**-6\n",
- "z2=3.3*10**-7\n",
- "t=900 #s\n",
- "e=12 #V\n",
- "\n",
- "#Calculation\n",
- "m2=(m1*z2)/z1\n",
- "d=(e*m1)/z1\n",
- "\n",
- "#Result\n",
- "print\"(a) Mass of the copper deposited is\",round(m2*10**3,3),\"10**-3\"\n",
- "print\"(b) The energy supplied by the battery is\",round(d*10**-3,2),\"10**3\",\"J\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a) Mass of the copper deposited is 0.119 10**-3\n",
- "(b) The energy supplied by the battery is 4.32 10**3 J\n"
- ]
- }
- ],
- "prompt_number": 51
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 9.6 Page no 286"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "r=100 #ohm\n",
- "t=600 #s\n",
- "z=3.3*10**-7\n",
- "m=10**-4\n",
- "\n",
- "#Calculation\n",
- "I=m/(z*t)\n",
- "Q=I**2*r*t\n",
- "\n",
- "#Result\n",
- "print\"Heat produced in the resistance coil is\",round(Q,1),\"J\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Heat produced in the resistance coil is 15304.6 J\n"
- ]
- }
- ],
- "prompt_number": 64
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 9.7 Page no 286"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "a=1.008\n",
- "v=1\n",
- "m1=1.05*10**-8\n",
- "a1=63.54\n",
- "v1=2\n",
- "m2=3.29*10**-7\n",
- "a2=107.9\n",
- "v2=1\n",
- "m3=1.12*10**-6\n",
- "a3=55.85\n",
- "v3=3\n",
- "\n",
- "#Calculation\n",
- "#For water voltameter\n",
- "E=a/v\n",
- "s=m1/E\n",
- "\n",
- "#For copper voltameter\n",
- "E2=a1/v1\n",
- "s1=m2/E2\n",
- "\n",
- "#For silver voltameter\n",
- "E3=a2/v2\n",
- "s2=m3/E3\n",
- "\n",
- "#For iron voltameter\n",
- "E4=a3/v3\n",
- "m4=(s/a)*E4\n",
- "\n",
- "#Result\n",
- "print\"Mass of hydrogen liberated is\",round(s*10**8,4)*10**-8\n",
- "print\"Mass of copper deposited is\",round(s1*10**8,4)*10**-8\n",
- "print\"Mass of silver deposited is\",round(s2*10**8,4)*10**-8\n",
- "print\"Mass of iron deposited is\", round(m4*10**7,2)*10**-7,\"Kg/C\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Mass of hydrogen liberated is 1.0417e-08\n",
- "Mass of copper deposited is 1.0356e-08\n",
- "Mass of silver deposited is 1.038e-08\n",
- "Mass of iron deposited is 1.92e-07 Kg/C\n"
- ]
- }
- ],
- "prompt_number": 104
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 9.8 Page no 287"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "m=2.5 #g\n",
- "I=10 #A\n",
- "F=96500 #C/mol\n",
- "m1=63.5\n",
- "n=2.0\n",
- "\n",
- "#Calculation\n",
- "E=m1/n\n",
- "t=m*F/(E*I)\n",
- "\n",
- "#Result\n",
- "print\"Time required is\",round(t,1),\"S\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Time required is 759.8 S\n"
- ]
- }
- ],
- "prompt_number": 110
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 9.9 Page no 287"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "m=16.43 #g\n",
- "t=4000 #s\n",
- "F=96485 #C/mol\n",
- "m1=63.54\n",
- "n=2.0\n",
- "I1=12.6 #A\n",
- "\n",
- "#Calculation\n",
- "E=m1/n\n",
- "I=m*F/(E*t)\n",
- "I2=I1-I\n",
- "\n",
- "#Result\n",
- "print\"Error in the ammeter reading is\", round(I2,3),\"A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Error in the ammeter reading is 0.126 A\n"
- ]
- }
- ],
- "prompt_number": 119
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 9.10 Page no 287"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "t=600 #S\n",
- "m=5.92 #g\n",
- "F=96500 #C/mol\n",
- "V1=1.62 #V\n",
- "V2=1.34\n",
- "m1=63.5\n",
- "n=2.0\n",
- "\n",
- "#Calculation\n",
- "V=V1-V2\n",
- "E=m1/n\n",
- "I=m*F/(E*t)\n",
- "R=V/I\n",
- "\n",
- "#Result\n",
- "print\"Resistance of the voltmeter is\",round(R*10**3,2),\"m ohm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Resistance of the voltmeter is 9.34 m ohm\n"
- ]
- }
- ],
- "prompt_number": 123
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 9.11 Page no 287"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "E=8 #V\n",
- "r=1 #ohm\n",
- "R=15 #ohm\n",
- "E1=120\n",
- "t=300 #s\n",
- "\n",
- "#Calculation\n",
- "I=(E1-E)/(R+r)\n",
- "V=E+(I*r)\n",
- "E12=E*I*t\n",
- "\n",
- "#Result\n",
- "print\"(a) Current in the circuit is\", I,\"A\"\n",
- "print\"(b) Terminal voltage across the battery is\",V,\"V\"\n",
- "print\"(c) Chemical energy stored in the battery is\",E12,\"J\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a) Current in the circuit is 7 A\n",
- "(b) Terminal voltage across the battery is 15 V\n",
- "(c) Chemical energy stored in the battery is 16800 J\n"
- ]
- }
- ],
- "prompt_number": 133
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 9.12 Page no 288"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "I=10 #A\n",
- "t=300 #S\n",
- "m=2.016\n",
- "n=2.0\n",
- "n1=1.0\n",
- "m1=1.008\n",
- "F=96500\n",
- "V=22.4\n",
- "\n",
- "#Calculation\n",
- "q=I*t\n",
- "M=m/n\n",
- "M2=m1/n1\n",
- "q1=F*m/m1\n",
- "V1=V*q/q1\n",
- "\n",
- "#Result\n",
- "print\"Volume of hydrogen is\",round(V1,4),\"litre\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Volume of hydrogen is 0.3482 litre\n"
- ]
- }
- ],
- "prompt_number": 136
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 9.13 Page no 288"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "w=107.9 #g/mol\n",
- "r=2 #ohm\n",
- "E=12 #V\n",
- "V=10\n",
- "F=96500\n",
- "t=1800\n",
- "\n",
- "#Calculation\n",
- "R=r*(V/(E-V))\n",
- "I=E/(R+r)\n",
- "E=w/I\n",
- "z=E/F\n",
- "m=z*I*t\n",
- "\n",
- "#Result\n",
- "print\"Silver deposited at the cathode is\", round(m,2),\"g\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Silver deposited at the cathode is 2.01 g\n"
- ]
- }
- ],
- "prompt_number": 143
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 9.14 Page no 288"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "I=5.0 #A\n",
- "a=0.5 #mole\n",
- "n=1\n",
- "F=96500\n",
- "I1=10 #A\n",
- "t1=9650*2\n",
- "n1=2.0\n",
- "m=63.54\n",
- "m2=55.85\n",
- "n2=3.0\n",
- "\n",
- "#Calculation\n",
- "q=F*a\n",
- "t=q/I\n",
- "E=m/n1\n",
- "m1=(E*I1*t1)/F\n",
- "E3=m2/n2\n",
- "m3=(E3*I1*t1)/F\n",
- "\n",
- "#Result\n",
- "print\"Molar mass of copper is\", m1,\"equal to its atomic mass i.e 1 mole of copper is liberated.\"\n",
- "print\"Molar mass of iron is\",round(m3,3),\"Hence 2/3 mole of iron will be deposited.\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Molar mass of copper is 63.54 equal to its atomic mass i.e 1 mole of copper is liberated.\n",
- "Molar mass of iron is 37.233 Hence 2/3 mole of iron will be deposited.\n"
- ]
- }
- ],
- "prompt_number": 153
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/modern_physics_by_Satish_K._Gupta/chap_29.ipynb b/modern_physics_by_Satish_K._Gupta/chap_29.ipynb deleted file mode 100755 index 581d372d..00000000 --- a/modern_physics_by_Satish_K._Gupta/chap_29.ipynb +++ /dev/null @@ -1,298 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:15f32b3e96d10d200143bd7ffbc8811afc0974b65cf155210561266d411e6511"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 29 Wave Nature Of Matter"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 29.1 Page no 815"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "h=6.62*10**-34 #Js\n",
- "Me=9.1*10**-31 #Kg\n",
- "Ve=10**5 #m s**-1\n",
- "Mp=1.67*10**-27 #Kg\n",
- "Vp=10**5 #m s**-1\n",
- "\n",
- "#Calculation\n",
- "Le=h/(Me*Ve)\n",
- "Lp=h/(Mp*Vp)\n",
- "\n",
- "#Result\n",
- "print\"The de-broglie wavelength of electron is\",round(Le*10**9,2),\"*10**-9 m\"\n",
- "print\"The de-broglie wavelenght of proton is\",round(Lp*10**12,2),\"*10**-12 m\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The de-broglie wavelength of electron is 7.27 10**-9 m\n",
- "The de-broglie wavelenght of proton is 3.96 10**-12 m\n"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 29.2 Page no 816"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "M0=9.1*10**-31 #Kg\n",
- "h=6.62*10**-34 #J s\n",
- "V=0.5 #c\n",
- "V1=1.5*10**8\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "L=(h*math.sqrt(1-V**2))/(M0*V1)\n",
- "\n",
- "#Result\n",
- "print\"The de-broglie wavelenght is\",round(L*10**12,1),\"*10**-12 m\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The de-broglie wavelenght is 4.2 10**-12 m\n"
- ]
- }
- ],
- "prompt_number": 16
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 29.3 Page no 816"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "h=6.6*10**-34 #J s\n",
- "m=9.1*10**-31 #kg\n",
- "Iev=1.6*10**-19 #J\n",
- "E=6.4*10**-17 #J\n",
- "\n",
- "#Calculation\n",
- "L=h/(math.sqrt(2*m*E))\n",
- "\n",
- "#Result\n",
- "print\"The de-broglie wavelenght of electron is\",round(L*10**10,2),\"A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The de-broglie wavelenght of electron is 0.61 A\n"
- ]
- }
- ],
- "prompt_number": 23
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 29.4 Page no 816"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "T=300 #K\n",
- "K=1.38*10**-23 #J\n",
- "h=6.62*10**-34\n",
- "m=1.675*10**-27\n",
- "\n",
- "#Calculation\n",
- "E=(3/2.0)*K*T\n",
- "L=h/(math.sqrt(2*m*E))\n",
- "\n",
- "#Result\n",
- "print\"The de-broglie wavelenght associated with thermal neutrons is\",round(L*10**10,3),\"*10**-10 m\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The de-broglie wavelenght associated with thermal neutrons is 1.451 10**-10 m\n"
- ]
- }
- ],
- "prompt_number": 30
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 29.5 Page no 816"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "L=10**-10 #m\n",
- "h=6.62*10**-34\n",
- "c=3*10**8\n",
- "S=1.6*10**-19\n",
- "m=9.11*10**-31 #kg\n",
- "\n",
- "#Calculation\n",
- "#For X-ray photon of wavelength\n",
- "E=(h*c)/L\n",
- "E1=E/S\n",
- "#For electron of wavelength \n",
- "Mv=h/L\n",
- "E2=(Mv**2)/(2*m)\n",
- "E3=E2/S\n",
- "\n",
- "#Result\n",
- "print\"Energy of electron is\",round(E3,1),\"eV\"\n",
- "print\"It follows that X-ray photon has greater energy\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Energy of electron is 150.3 eV\n",
- "It follows that X-ray photon has greater energy\n"
- ]
- }
- ],
- "prompt_number": 42
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 29.6 Page no 816"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "M=1.675*10**-27 #kg\n",
- "E=2.4*10**-17 #J\n",
- "h=6.62*10**-34\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "L=h/(math.sqrt(2*M*E))\n",
- "\n",
- "#Result\n",
- "print\"The de-broglie wavelength of neutron is\",round(L*10**12,3)*10**-12,\" m\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The de-broglie wavelength of neutron is 2.335e-12 m\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 29.7 Page no 816"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Given\n",
- "V=50 #kV\n",
- "E=8.0*10**-15 #J\n",
- "m=9.1*10**-31\n",
- "h=6.62*10**-34\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "L=h/(math.sqrt(2*m*E))\n",
- "\n",
- "#Result\n",
- "print\"Energy of electron is\",round(L*10**12,3),\"*10**-12 m\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Energy of electron is 5.486 10**-12 m\n"
- ]
- }
- ],
- "prompt_number": 54
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file |