removing problem statements

1 files changed, 405 insertions, 381 deletions

diff --git a/Industrial_Instrumentation/Chapter_6.ipynb b/Industrial_Instrumentation/Chapter_6.ipynb
index d3a9a093..bdc5920d 100644
--- a/Industrial_Instrumentation/Chapter_6.ipynb
+++ b/Industrial_Instrumentation/Chapter_6.ipynb
@@ -1,534 +1,558 @@
 {
  "metadata": {
-  "name": "Chapter_6"
- }, 
- "nbformat": 2, 
+  "name": "",
+  "signature": "sha256:6148d2be2796832e7d2e9cefcc5c361a4c9dc07b22c355b627a74bd914199046"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
  "worksheets": [
   {
    "cells": [
     {
-     "cell_type": "markdown", 
+     "cell_type": "markdown",
+     "metadata": {},
      "source": [
       "<h1>Chapter 6: Level<h1>"
      ]
-    }, 
+    },
     {
-     "cell_type": "markdown", 
+     "cell_type": "markdown",
+     "metadata": {},
      "source": [
       "<h3>Example 6.1,Page Number:370<h3>"
      ]
-    }, 
+    },
     {
-     "cell_type": "code", 
-     "collapsed": false, 
+     "cell_type": "code",
+     "collapsed": false,
      "input": [
-      "'''output current of two wire pressure transmitter'''", 
-      "", 
-      "#(a)", 
-      "", 
-      "# variable declaration", 
-      "p=1.5                      # pressure applied", 
-      "a=4.0                      # mA corresponds to 0 kg/cm^2", 
-      "b=20.0                     # mA corresponds to 2 kg/cm^2", 
-      "", 
-      "#calculation", 
-      "wh=(((b-a)/2)*p)+a", 
-      "", 
-      "#result", 
-      "print('(a)just at the bottom level of the tank')", 
-      "print('Water head applied to the transmitter =%d mA'%wh)", 
-      "", 
-      "#(b)", 
-      "", 
-      "#calculation", 
-      "wh2=(((b-a)/2)*p)+2*a", 
-      "", 
-      "#result", 
-      "print('\\n\\n(b)5m below the bottom of the tank')", 
-      "print('Water head applied to the transmitter =%d mA' %wh2)", 
-      "", 
-      "#(c)", 
-      "", 
-      "#calculation", 
-      "wh3=(((b-a)/2)*p)", 
-      "", 
-      "#result", 
-      "print('\\n\\n(c)5m above the bottom of the tank')", 
+      "\n",
+      "\n",
+      "#(a)\n",
+      "\n",
+      "# variable declaration\n",
+      "p=1.5                      # pressure applied\n",
+      "a=4.0                      # mA corresponds to 0 kg/cm^2\n",
+      "b=20.0                     # mA corresponds to 2 kg/cm^2\n",
+      "\n",
+      "#calculation\n",
+      "wh=(((b-a)/2)*p)+a\n",
+      "\n",
+      "#result\n",
+      "print('(a)just at the bottom level of the tank')\n",
+      "print('Water head applied to the transmitter =%d mA'%wh)\n",
+      "\n",
+      "#(b)\n",
+      "\n",
+      "#calculation\n",
+      "wh2=(((b-a)/2)*p)+2*a\n",
+      "\n",
+      "#result\n",
+      "print('\\n\\n(b)5m below the bottom of the tank')\n",
+      "print('Water head applied to the transmitter =%d mA' %wh2)\n",
+      "\n",
+      "#(c)\n",
+      "\n",
+      "#calculation\n",
+      "wh3=(((b-a)/2)*p)\n",
+      "\n",
+      "#result\n",
+      "print('\\n\\n(c)5m above the bottom of the tank')\n",
       "print('Water head applied to the transmitter =%d mA'%wh3)"
-     ], 
-     "language": "python", 
+     ],
+     "language": "python",
+     "metadata": {},
      "outputs": [
       {
-       "output_type": "stream", 
-       "stream": "stdout", 
+       "output_type": "stream",
+       "stream": "stdout",
        "text": [
-        "(a)just at the bottom level of the tank", 
-        "Water head applied to the transmitter =16 mA", 
-        "", 
-        "", 
-        "(b)5m below the bottom of the tank", 
-        "Water head applied to the transmitter =20 mA", 
-        "", 
-        "", 
-        "(c)5m above the bottom of the tank", 
+        "(a)just at the bottom level of the tank\n",
+        "Water head applied to the transmitter =16 mA\n",
+        "\n",
+        "\n",
+        "(b)5m below the bottom of the tank\n",
+        "Water head applied to the transmitter =20 mA\n",
+        "\n",
+        "\n",
+        "(c)5m above the bottom of the tank\n",
         "Water head applied to the transmitter =12 mA"
        ]
       }
-     ], 
+     ],
      "prompt_number": 1
-    }, 
+    },
     {
-     "cell_type": "markdown", 
+     "cell_type": "markdown",
+     "metadata": {},
      "source": [
       "<h3>Example 6.2, Page Number:371<h3>"
      ]
-    }, 
+    },
     {
-     "cell_type": "code", 
-     "collapsed": false, 
+     "cell_type": "code",
+     "collapsed": false,
      "input": [
-      "'''water level and current at different positions'''", 
-      "", 
-      "#(a)", 
-      "", 
-      "#variable declaration", 
-      "b=20.0             # Maximum output", 
-      "a=4.0              # minimum output ", 
-      "op=16.0            # output in mA", 
-      "", 
-      "#calculation", 
-      "p=(op-a)*2/(b-a)", 
-      "p_h=p*10.0", 
-      "h=p_h-2-5", 
-      "", 
-      "#result", 
-      "print('(a)\\nh = %dm'%h)", 
-      "", 
-      "#(b)", 
-      "", 
-      "#variable declaration", 
-      "p1=1                # pressure applied", 
-      "", 
-      "#calculation", 
-      "t_op=((b-a)/2)*p1+4", 
-      "", 
-      "#result", 
-      "print('\\n(b)\\nTransmitter output =%d mA'%t_op)", 
-      "", 
-      "#(c)", 
-      "", 
-      "#variable declaration", 
-      "p2=0.5                    # applied pressure", 
-      "", 
-      "#calculation", 
-      "t_op1=((b-a)/2)*p2+4", 
-      "", 
-      "#result", 
+      "\n",
+      "\n",
+      "#(a)\n",
+      "\n",
+      "#variable declaration\n",
+      "b=20.0             # Maximum output\n",
+      "a=4.0              # minimum output \n",
+      "op=16.0            # output in mA\n",
+      "\n",
+      "#calculation\n",
+      "p=(op-a)*2/(b-a)\n",
+      "p_h=p*10.0\n",
+      "h=p_h-2-5\n",
+      "\n",
+      "#result\n",
+      "print('(a)\\nh = %dm'%h)\n",
+      "\n",
+      "#(b)\n",
+      "\n",
+      "#variable declaration\n",
+      "p1=1                # pressure applied\n",
+      "\n",
+      "#calculation\n",
+      "t_op=((b-a)/2)*p1+4\n",
+      "\n",
+      "#result\n",
+      "print('\\n(b)\\nTransmitter output =%d mA'%t_op)\n",
+      "\n",
+      "#(c)\n",
+      "\n",
+      "#variable declaration\n",
+      "p2=0.5                    # applied pressure\n",
+      "\n",
+      "#calculation\n",
+      "t_op1=((b-a)/2)*p2+4\n",
+      "\n",
+      "#result\n",
       "print('\\n(c)\\nTransmitter output =%d mA'%t_op1)"
-     ], 
-     "language": "python", 
+     ],
+     "language": "python",
+     "metadata": {},
      "outputs": [
       {
-       "output_type": "stream", 
-       "stream": "stdout", 
+       "output_type": "stream",
+       "stream": "stdout",
        "text": [
-        "(a)", 
-        "h = 8m", 
-        "", 
-        "(b)", 
-        "Transmitter output =12 mA", 
-        "", 
-        "(c)", 
+        "(a)\n",
+        "h = 8m\n",
+        "\n",
+        "(b)\n",
+        "Transmitter output =12 mA\n",
+        "\n",
+        "(c)\n",
         "Transmitter output =8 mA"
        ]
       }
-     ], 
+     ],
      "prompt_number": 2
-    }, 
+    },
     {
-     "cell_type": "markdown", 
+     "cell_type": "markdown",
+     "metadata": {},
      "source": [
       "<h3>Example 6.3, Page Number: 372<h3>"
      ]
-    }, 
+    },
     {
-     "cell_type": "code", 
-     "collapsed": false, 
+     "cell_type": "code",
+     "collapsed": false,
      "input": [
-      "'''Differential pressure output at different levels'''", 
-      "", 
-      "#(a)", 
-      "", 
-      "#variable declaration", 
-      "b=20.0                  # Maximum output", 
-      "a=4.0                   # minimum output", 
-      "op=16.0                 # actual output   ", 
-      "wt_l1=25.0              # water level (i)", 
-      "", 
-      "#calculation", 
-      "t_op=((b-a)/100)*(100-75)+4", 
-      "", 
-      "#result", 
-      "print('(a)\\nWater level=+25cm\\nTransmitter output = %d mA' %t_op)", 
-      "", 
-      "#(b)", 
-      "", 
-      "#calculation", 
-      "wt_l2=-25.0            # water level (ii)", 
-      "t_op2=((b-a)/100)*(100-25)+4", 
-      "", 
-      "#result", 
-      "print('\\n(b)\\nWater level=-25cm\\nTransmitter output = %d mA' %t_op2)", 
-      "", 
-      "#(c)", 
-      "", 
-      "#Variable declaration", 
-      "t_op3=12.0             # Transmitter output ", 
-      "", 
-      "#calculation", 
-      "H=(100.0/(b-a))*(12-4) ", 
-      "", 
-      "#result", 
+      "\n",
+      "\n",
+      "#(a)\n",
+      "\n",
+      "#variable declaration\n",
+      "b=20.0                  # Maximum output\n",
+      "a=4.0                   # minimum output\n",
+      "op=16.0                 # actual output   \n",
+      "wt_l1=25.0              # water level (i)\n",
+      "\n",
+      "#calculation\n",
+      "t_op=((b-a)/100)*(100-75)+4\n",
+      "\n",
+      "#result\n",
+      "print('(a)\\nWater level=+25cm\\nTransmitter output = %d mA' %t_op)\n",
+      "\n",
+      "#(b)\n",
+      "\n",
+      "#calculation\n",
+      "wt_l2=-25.0            # water level (ii)\n",
+      "t_op2=((b-a)/100)*(100-25)+4\n",
+      "\n",
+      "#result\n",
+      "print('\\n(b)\\nWater level=-25cm\\nTransmitter output = %d mA' %t_op2)\n",
+      "\n",
+      "#(c)\n",
+      "\n",
+      "#Variable declaration\n",
+      "t_op3=12.0             # Transmitter output \n",
+      "\n",
+      "#calculation\n",
+      "H=(100.0/(b-a))*(12-4) \n",
+      "\n",
+      "#result\n",
       "print('\\n(c)\\nHead Applied = %d cm\\nLevel corresponding to 50 cm head =0 cm' %H)"
-     ], 
-     "language": "python", 
+     ],
+     "language": "python",
+     "metadata": {},
      "outputs": [
       {
-       "output_type": "stream", 
-       "stream": "stdout", 
+       "output_type": "stream",
+       "stream": "stdout",
        "text": [
-        "(a)", 
-        "Water level=+25cm", 
-        "Transmitter output = 8 mA", 
-        "", 
-        "(b)", 
-        "Water level=-25cm", 
-        "Transmitter output = 16 mA", 
-        "", 
-        "(c)", 
-        "Head Applied = 50 cm", 
+        "(a)\n",
+        "Water level=+25cm\n",
+        "Transmitter output = 8 mA\n",
+        "\n",
+        "(b)\n",
+        "Water level=-25cm\n",
+        "Transmitter output = 16 mA\n",
+        "\n",
+        "(c)\n",
+        "Head Applied = 50 cm\n",
         "Level corresponding to 50 cm head =0 cm"
        ]
       }
-     ], 
+     ],
      "prompt_number": 3
-    }, 
+    },
     {
-     "cell_type": "markdown", 
+     "cell_type": "markdown",
+     "metadata": {},
      "source": [
       "<h3>Example 6.4, Page Number: 373<h3>"
      ]
-    }, 
+    },
     {
-     "cell_type": "code", 
-     "collapsed": false, 
+     "cell_type": "code",
+     "collapsed": false,
      "input": [
-      "'''Displacer with spring balance'''", 
-      "", 
-      "#(a)", 
-      "", 
-      "#variable declaration", 
-      "a=5.0*10**-4                  #area", 
-      "l=8.0                         #length", 
-      "dens=6.0*1000.0               #density", 
-      "", 
-      "#calculation", 
-      "w=a*l*dens", 
-      "", 
-      "#result", 
-      "print('(a)\\nWeight of the displacer if weighed in air = %d kg'%w)", 
-      "", 
-      "", 
-      "#(i)", 
-      "", 
-      "#variable declaration", 
-      "sbr1=23.0                   # spring balance reading", 
-      "", 
-      "#calculation", 
-      "wloss1=w-sbr1", 
-      "L1=wloss1/(1000.0*a)", 
-      "", 
-      "#result", 
-      "print('\\n(i)\\tL1=%dm'%L1)", 
-      "", 
-      "", 
-      "#(ii)", 
-      "", 
-      "#variable declaration", 
-      "sbr2=22.0                  # spring balance reading", 
-      "", 
-      "#calculation", 
-      "wloss2=w-sbr2", 
-      "L2=wloss2/(1000.0*a)", 
-      "", 
-      "#result", 
-      "print('\\n(ii)\\tL2=%dm'%L2)", 
-      "", 
-      "#(iii)", 
-      "", 
-      "#variable declaration", 
-      "sbr3=21.0                # spring balance reading", 
-      "", 
-      "#calculation", 
-      "wloss3=w-sbr3", 
-      "L3=wloss3/(1000.0*a)", 
-      "", 
-      "#result", 
-      "print('\\n(iii)\\tL3=%dm'%L3)", 
-      "", 
-      "#(b)", 
-      "", 
-      "#variable declaration", 
-      "level=8.0                # level wen tank is full     ", 
-      "", 
-      "#calculation", 
-      "wt=a*level*1000.0", 
-      "spring=w-wt", 
-      "", 
-      "#result", 
+      "\n",
+      "\n",
+      "#(a)\n",
+      "\n",
+      "#variable declaration\n",
+      "a=5.0*10**-4                  #area\n",
+      "l=8.0                         #length\n",
+      "dens=6.0*1000.0               #density\n",
+      "\n",
+      "#calculation\n",
+      "w=a*l*dens\n",
+      "\n",
+      "#result\n",
+      "print('(a)\\nWeight of the displacer if weighed in air = %d kg'%w)\n",
+      "\n",
+      "\n",
+      "#(i)\n",
+      "\n",
+      "#variable declaration\n",
+      "sbr1=23.0                   # spring balance reading\n",
+      "\n",
+      "#calculation\n",
+      "wloss1=w-sbr1\n",
+      "L1=wloss1/(1000.0*a)\n",
+      "\n",
+      "#result\n",
+      "print('\\n(i)\\tL1=%dm'%L1)\n",
+      "\n",
+      "\n",
+      "#(ii)\n",
+      "\n",
+      "#variable declaration\n",
+      "sbr2=22.0                  # spring balance reading\n",
+      "\n",
+      "#calculation\n",
+      "wloss2=w-sbr2\n",
+      "L2=wloss2/(1000.0*a)\n",
+      "\n",
+      "#result\n",
+      "print('\\n(ii)\\tL2=%dm'%L2)\n",
+      "\n",
+      "#(iii)\n",
+      "\n",
+      "#variable declaration\n",
+      "sbr3=21.0                # spring balance reading\n",
+      "\n",
+      "#calculation\n",
+      "wloss3=w-sbr3\n",
+      "L3=wloss3/(1000.0*a)\n",
+      "\n",
+      "#result\n",
+      "print('\\n(iii)\\tL3=%dm'%L3)\n",
+      "\n",
+      "#(b)\n",
+      "\n",
+      "#variable declaration\n",
+      "level=8.0                # level wen tank is full     \n",
+      "\n",
+      "#calculation\n",
+      "wt=a*level*1000.0\n",
+      "spring=w-wt\n",
+      "\n",
+      "#result\n",
       "print('\\n(b):when the tank is full\\nSpring Balance reading = %d kg'%spring)"
-     ], 
-     "language": "python", 
+     ],
+     "language": "python",
+     "metadata": {},
      "outputs": [
       {
-       "output_type": "stream", 
-       "stream": "stdout", 
+       "output_type": "stream",
+       "stream": "stdout",
        "text": [
-        "(a)", 
-        "Weight of the displacer if weighed in air = 24 kg", 
-        "", 
-        "(i)\tL1=2m", 
-        "", 
-        "(ii)\tL2=4m", 
-        "", 
-        "(iii)\tL3=6m", 
-        "", 
-        "(b):when the tank is full", 
+        "(a)\n",
+        "Weight of the displacer if weighed in air = 24 kg\n",
+        "\n",
+        "(i)\tL1=2m\n",
+        "\n",
+        "(ii)\tL2=4m\n",
+        "\n",
+        "(iii)\tL3=6m\n",
+        "\n",
+        "(b):when the tank is full\n",
         "Spring Balance reading = 20 kg"
        ]
       }
-     ], 
+     ],
      "prompt_number": 4
-    }, 
+    },
     {
-     "cell_type": "markdown", 
+     "cell_type": "markdown",
+     "metadata": {},
      "source": [
       "<h3>Example 6.5, Page Number: 374<h3>"
      ]
-    }, 
+    },
     {
-     "cell_type": "code", 
-     "collapsed": false, 
+     "cell_type": "code",
+     "collapsed": false,
      "input": [
-      "'''Buoyancy Force calculation'''", 
-      "", 
-      "#variable declaration", 
-      "rho=1000.0                # density of water ", 
-      "v=3.0                     # displaced volume of water ", 
-      "", 
-      "#calculation", 
-      "Bw=rho*v", 
-      "", 
-      "#Result", 
+      "\n",
+      "\n",
+      "#variable declaration\n",
+      "rho=1000.0                # density of water \n",
+      "v=3.0                     # displaced volume of water \n",
+      "\n",
+      "#calculation\n",
+      "Bw=rho*v\n",
+      "\n",
+      "#Result\n",
       "print('Buoyance Force(Bw) = %d kg'%Bw)"
-     ], 
-     "language": "python", 
+     ],
+     "language": "python",
+     "metadata": {},
      "outputs": [
       {
-       "output_type": "stream", 
-       "stream": "stdout", 
+       "output_type": "stream",
+       "stream": "stdout",
        "text": [
         "Buoyance Force(Bw) = 3000 kg"
        ]
       }
-     ], 
+     ],
      "prompt_number": 5
-    }, 
+    },
     {
-     "cell_type": "markdown", 
+     "cell_type": "markdown",
+     "metadata": {},
      "source": [
       "<h3>Example 6.6, Page Number: 374<h3>"
      ]
-    }, 
+    },
     {
-     "cell_type": "code", 
-     "collapsed": false, 
+     "cell_type": "code",
+     "collapsed": false,
      "input": [
-      "'''Determination of displaced volume from Buoyancy Force'''", 
-      "", 
-      "#variable declaration", 
-      "rho=1000.0             # density of water", 
-      "Bw=5000.0              # Buoyancy  Force", 
-      "", 
-      "#calculation", 
-      "v=Bw/rho", 
-      "", 
-      "#result", 
+      "\n",
+      "\n",
+      "#variable declaration\n",
+      "rho=1000.0             # density of water\n",
+      "Bw=5000.0              # Buoyancy  Force\n",
+      "\n",
+      "#calculation\n",
+      "v=Bw/rho\n",
+      "\n",
+      "#result\n",
       "print('V = %d m^3' %v)"
-     ], 
-     "language": "python", 
+     ],
+     "language": "python",
+     "metadata": {},
      "outputs": [
       {
-       "output_type": "stream", 
-       "stream": "stdout", 
+       "output_type": "stream",
+       "stream": "stdout",
        "text": [
         "V = 5 m^3"
        ]
       }
-     ], 
+     ],
      "prompt_number": 6
-    }, 
+    },
     {
-     "cell_type": "markdown", 
+     "cell_type": "markdown",
+     "metadata": {},
      "source": [
       "<h3>Example 6.7, Page Number: 374<h3>"
      ]
-    }, 
+    },
     {
-     "cell_type": "code", 
-     "collapsed": false, 
+     "cell_type": "code",
+     "collapsed": false,
      "input": [
-      "'''Determination of hydrostatic pressure in open tank'''", 
-      "", 
-      "#variable declaration", 
-      "rho=1000.0           # density of water", 
-      "h=10.0               # height of liquid", 
-      "", 
-      "#calculation", 
-      "P=rho*h", 
-      "", 
-      "#result", 
+      "\n",
+      "\n",
+      "#variable declaration\n",
+      "rho=1000.0           # density of water\n",
+      "h=10.0               # height of liquid\n",
+      "\n",
+      "#calculation\n",
+      "P=rho*h\n",
+      "\n",
+      "#result\n",
       "print('P = %d kg/m^2 = %d kg/cm^2 '%(P,P/10000))"
-     ], 
-     "language": "python", 
+     ],
+     "language": "python",
+     "metadata": {},
      "outputs": [
       {
-       "output_type": "stream", 
-       "stream": "stdout", 
+       "output_type": "stream",
+       "stream": "stdout",
        "text": [
         "P = 10000 kg/m^2 = 1 kg/cm^2 "
        ]
       }
-     ], 
+     ],
      "prompt_number": 7
-    }, 
+    },
     {
-     "cell_type": "markdown", 
+     "cell_type": "markdown",
+     "metadata": {},
      "source": [
       "<h3>Example 6.8, Page Number: 374<h3>"
      ]
-    }, 
+    },
     {
-     "cell_type": "code", 
-     "collapsed": false, 
+     "cell_type": "code",
+     "collapsed": false,
      "input": [
-      "'''Determination of hydrostatic pressure in closed tank'''", 
-      "", 
-      "#variable declaration", 
-      "rho=1000.0               # density of water", 
-      "h=15.0                   # height of liquid ", 
-      "ex_p=1.0                 # External pressure on liquid", 
-      "", 
-      "#calculation", 
-      "P=(rho*h/10000.0)+ex_p", 
-      "", 
-      "#result", 
+      "\n",
+      "\n",
+      "#variable declaration\n",
+      "rho=1000.0               # density of water\n",
+      "h=15.0                   # height of liquid \n",
+      "ex_p=1.0                 # External pressure on liquid\n",
+      "\n",
+      "#calculation\n",
+      "P=(rho*h/10000.0)+ex_p\n",
+      "\n",
+      "#result\n",
       "print('P = %.1f kg/cm^2' %P)"
-     ], 
-     "language": "python", 
+     ],
+     "language": "python",
+     "metadata": {},
      "outputs": [
       {
-       "output_type": "stream", 
-       "stream": "stdout", 
+       "output_type": "stream",
+       "stream": "stdout",
        "text": [
         "P = 2.5 kg/cm^2"
        ]
       }
-     ], 
+     ],
      "prompt_number": 8
-    }, 
+    },
     {
-     "cell_type": "markdown", 
+     "cell_type": "markdown",
+     "metadata": {},
      "source": [
       "<h3>Example 6.9, Page Number:374<h3>"
      ]
-    }, 
+    },
     {
-     "cell_type": "code", 
-     "collapsed": false, 
+     "cell_type": "code",
+     "collapsed": false,
      "input": [
-      "'''Determination of height from hydrostatic pressure'''", 
-      "", 
-      "#variable declaration", 
-      "rho=1000.0               # density of water", 
-      "ex_p=0.5*10**4           # External pressure on liquid     ", 
-      "P=1.6*10**4              #(rho*h/10000)+ex_p", 
-      "", 
-      "#calculation", 
-      "h=(P-ex_p)/1000.0", 
-      "", 
-      "#result", 
+      "\n",
+      "\n",
+      "#variable declaration\n",
+      "rho=1000.0               # density of water\n",
+      "ex_p=0.5*10**4           # External pressure on liquid     \n",
+      "P=1.6*10**4              #(rho*h/10000)+ex_p\n",
+      "\n",
+      "#calculation\n",
+      "h=(P-ex_p)/1000.0\n",
+      "\n",
+      "#result\n",
       "print('h = %d m' %h)"
-     ], 
-     "language": "python", 
+     ],
+     "language": "python",
+     "metadata": {},
      "outputs": [
       {
-       "output_type": "stream", 
-       "stream": "stdout", 
+       "output_type": "stream",
+       "stream": "stdout",
        "text": [
         "h = 11 m"
        ]
       }
-     ], 
+     ],
      "prompt_number": 9
-    }, 
+    },
     {
-     "cell_type": "markdown", 
+     "cell_type": "markdown",
+     "metadata": {},
      "source": [
       "<h3>Example 6.10, Page Number:375<h3>"
      ]
-    }, 
+    },
     {
-     "cell_type": "code", 
-     "collapsed": false, 
+     "cell_type": "code",
+     "collapsed": false,
      "input": [
-      "'''calculation of level on the probe'''", 
-      "", 
-      "#variable declaration", 
-      "c2=100.0*10**-6          # capacitance in capacitance probe", 
-      "r1=10.0*10**3            # value of resistor in bride", 
-      "r2=100.0*10**3           # value of resistor in bride", 
-      "r3=50.0*10**3            # value of resistor in bride", 
-      "", 
-      "#calculation", 
-      "Cx=r1*c2/r3", 
-      "Cx=Cx*10**6", 
-      "", 
-      "#result", 
-      "print('Cx = %d microFarad'%Cx)", 
-      "c=5.0", 
-      "", 
-      "#calculation", 
-      "l=Cx/c", 
-      "", 
-      "#result", 
+      "\n",
+      "\n",
+      "#variable declaration\n",
+      "c2=100.0*10**-6          # capacitance in capacitance probe\n",
+      "r1=10.0*10**3            # value of resistor in bride\n",
+      "r2=100.0*10**3           # value of resistor in bride\n",
+      "r3=50.0*10**3            # value of resistor in bride\n",
+      "\n",
+      "#calculation\n",
+      "Cx=r1*c2/r3\n",
+      "Cx=Cx*10**6\n",
+      "\n",
+      "#result\n",
+      "print('Cx = %d microFarad'%Cx)\n",
+      "c=5.0\n",
+      "\n",
+      "#calculation\n",
+      "l=Cx/c\n",
+      "\n",
+      "#result\n",
       "print('\\nLevel on the probe = %dm'%l)"
-     ], 
-     "language": "python", 
+     ],
+     "language": "python",
+     "metadata": {},
      "outputs": [
       {
-       "output_type": "stream", 
-       "stream": "stdout", 
+       "output_type": "stream",
+       "stream": "stdout",
        "text": [
-        "Cx = 20 microFarad", 
-        "", 
+        "Cx = 20 microFarad\n",
+        "\n",
         "Level on the probe = 4m"
        ]
       }
-     ], 
+     ],
      "prompt_number": 10
     }
-   ]
+   ],
+   "metadata": {}
   }
  ]
 }
\ No newline at end of file