{
 "metadata": {
  "name": "chapter8"
 },
 "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": "# To calculate the surface area to volume ratio\n\n#import modules\nimport math\nfrom __future__ import division\n\n#Variable decleration\nr=5;    #radius in m\npi=3.14;\n\n#Calculation \nSA=4*pi*r**2;    #surface area of sphere in m^2\nV=(4/3)*pi*r**3;   #volume of sphere in m^3\nR=SA/V;    #ratio\n#surface area to volume ratio can also be given by 3/radius\n\n#Result\nprint(\"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": "# To calculate the surface area to volume ratio\n\n#import modules\nimport math\nfrom __future__ import division\n\n#Variable decleration\nd=26;   #distance in m\nr=d/2;  #radius in m\npi=3.14;\n\n#Calculation\nSA=4*pi*r**2;      #surface area of sphere in m^2\nV=(4/3)*pi*r**3;   #volume of sphere in m^3\nR=SA/V;    #ratio\nR=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\nprint(\"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": "# To calculate the volume of cone\n\n#import modules\nimport math\nfrom __future__ import division\n\n#Variable decleration\nr=1;   #radius in m\nh=1;   #height in m\npi=3.14\n\n#Calculation\nV=(1/3)*pi*(r**2)*h;\nV=math.ceil(V*10**2)/10**2;   #rounding off to 2 decimals\n\n#Result\nprint(\"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": "# To calculate the total surface area of cone\n\n#import modules\nimport math\nfrom __future__ import division\n\n#Variable decleration\nr=3;   # radius in m\nh=4;   # height in m\npi=3.14\n\n#Calculation\nSA=pi*r*math.sqrt((r**2)+(h**2));\nTSA=SA+(pi*r**2);\n\n#Result\nprint(\"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": "# To calculate the height of cone\n\n#import modules\nimport math\nfrom __future__ import division\n\n#Variable decleration\nV=100;   #volume of cone in cubic inches\nr=5;   #radius of cone in inches\npi=3.14;\n\n#Calculation\nr_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)\nh=(3*V)/(pi*r**2);   #height in inches\nR=3/r_m;\nh=math.ceil(h*10**3)/10**3;   #rounding off to 3 decimals\n\n#Result\nprint(\"height of the cone in inches is\",h);\nprint(\"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": {}
  }
 ]
}