diff options
Diffstat (limited to 'modern_physics_by_Satish_K._Gupta/chap9.ipynb')
| -rwxr-xr-x | modern_physics_by_Satish_K._Gupta/chap9.ipynb | 627 |
1 files changed, 0 insertions, 627 deletions
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 |