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diff --git a/Engineering_Physics_by_Dr._K._Vijaya_Kumar/chapter8_2.ipynb b/Engineering_Physics_by_Dr._K._Vijaya_Kumar/chapter8_2.ipynb new file mode 100755 index 00000000..2dc13b1f --- /dev/null +++ b/Engineering_Physics_by_Dr._K._Vijaya_Kumar/chapter8_2.ipynb @@ -0,0 +1,253 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:064d55405a5d05f007b28f32cf39a9f99d10f303fc4084e2d14d99aaeb87858c" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Physics of Nano Materials" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example number 8.1, Page number 320" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "#import modules\n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable decleration\n", + "r=5; #radius in m\n", + "pi=3.14;\n", + "\n", + "#Calculation \n", + "SA=4*pi*r**2; #surface area of sphere in m^2\n", + "V=(4/3)*pi*r**3; #volume of sphere in m^3\n", + "R=SA/V; #ratio\n", + "#surface area to volume ratio can also be given by 3/radius\n", + "\n", + "#Result\n", + "print(\"surface area to volume ratio of sphere in m-1 is\",R);" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "('surface area to volume ratio of sphere in m-1 is', 0.6)\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example number 8.2, Page number 321" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "#import modules\n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable decleration\n", + "d=26; #distance in m\n", + "r=d/2; #radius in m\n", + "pi=3.14;\n", + "\n", + "#Calculation\n", + "SA=4*pi*r**2; #surface area of sphere in m^2\n", + "V=(4/3)*pi*r**3; #volume of sphere in m^3\n", + "R=SA/V; #ratio\n", + "R=math.ceil(R*10**3)/10**3; #rounding off to 3 decimals\n", + "#surface area to volume ratio can also be given by 3/radius\n", + "\n", + "#Result\n", + "print(\"surface area to volume ratio of sphere in m-1 is\",R);" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "('surface area to volume ratio of sphere in m-1 is', 0.231)\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example number 8.3, Page number 321" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "#import modules\n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable decleration\n", + "r=1; #radius in m\n", + "h=1; #height in m\n", + "pi=3.14\n", + "\n", + "#Calculation\n", + "V=(1/3)*pi*(r**2)*h;\n", + "V=math.ceil(V*10**2)/10**2; #rounding off to 2 decimals\n", + "\n", + "#Result\n", + "print(\"volume of cone in m^3 is\",V); " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "('volume of cone in m^3 is', 1.05)\n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example number 8.4, Page number 321" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "#import modules\n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable decleration\n", + "r=3; # radius in m\n", + "h=4; # height in m\n", + "pi=3.14\n", + "\n", + "#Calculation\n", + "SA=pi*r*math.sqrt((r**2)+(h**2));\n", + "TSA=SA+(pi*r**2);\n", + "\n", + "#Result\n", + "print(\"total surface area of cone in m^2 is\",TSA);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "('total surface area of cone in m^2 is', 75.36)\n" + ] + } + ], + "prompt_number": 16 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example number 8.5, Page number 322" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "#import modules\n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable decleration\n", + "V=100; #volume of cone in cubic inches\n", + "r=5; #radius of cone in inches\n", + "pi=3.14;\n", + "\n", + "#Calculation\n", + "r_m=r*0.0254; #radius of cone in m\n", + "#volume V=(1/3)*pi*(r**2)*h\n", + "#therefore h = (3*V)/(pi*r**2)\n", + "h=(3*V)/(pi*r**2); #height in inches\n", + "R=3/r_m;\n", + "h=math.ceil(h*10**3)/10**3; #rounding off to 3 decimals\n", + "\n", + "#Result\n", + "print(\"height of the cone in inches is\",h);\n", + "print(\"surface area to volume ratio in m-1 is\",R);\n", + "\n", + "#answer for the surface area to volume ratio given in the book is wrong" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "('height of the cone in inches is', 3.822)\n", + "('surface area to volume ratio in m-1 is', 23.62204724409449)\n" + ] + } + ], + "prompt_number": 18 + }, + { + "cell_type": "code", + "collapsed": false, + "input": [], + "language": "python", + "metadata": {}, + "outputs": [] + } + ], + "metadata": {} + } + ] +}
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