Added(A)/Deleted(D) following books

 A  1000_solved_Problems_in_Fluid_Mechanics_includes_Hydraulic_machines_by_K.Subramanya/README.txt
A  Fundamentals_Of_Aerodynamics_by_J._D._Anderson_Jr./README.txt
A  Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/README.txt
A  SURVYNG_AND_LEVELLING__by_N.N.BASAK/README.txt
A  sample_notebooks/Harshitgarg/Chapter_1-INTRODUCTION_TO_MECHANICS_OF_SOLIDS__2.ipynb
A  "sample_notebooks/KARTHIKEYAN S/CHAPTER_1.ipynb"
A  sample_notebooks/NityaL/Sample-Chapter_26.ipynb

3 files changed, 710 insertions, 0 deletions

diff --git a/sample_notebooks/Harshitgarg/Chapter_1-INTRODUCTION_TO_MECHANICS_OF_SOLIDS__2.ipynb b/sample_notebooks/Harshitgarg/Chapter_1-INTRODUCTION_TO_MECHANICS_OF_SOLIDS__2.ipynb
new file mode 100644
index 00000000..696f0022
--- /dev/null
+++ b/sample_notebooks/Harshitgarg/Chapter_1-INTRODUCTION_TO_MECHANICS_OF_SOLIDS__2.ipynb
@@ -0,0 +1,380 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {
+    "collapsed": true
+   },
+   "source": [
+    "# example1.1 Page number 10\n"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      " The resultant velocity : 21.54 km/hour\n",
+      "68.2 °\n"
+     ]
+    }
+   ],
+   "source": [
+    "#downstream direction as x\n",
+    "#direction across river as y\n",
+    "\n",
+    "from math import sqrt,atan,pi\n",
+    "\n",
+    "#variable declaration\n",
+    "\n",
+    "Vx= 8                       #velocity of stream, km/hour\n",
+    "Vy=float(20)                       #velocity of boat,km/hour\n",
+    "\n",
+    "V=sqrt(pow(Vx,2)+pow(Vy,2)) #resultant velocity, km/hour\n",
+    "theta=Vy/Vx\n",
+    "\n",
+    "alpha= atan(theta)*180/pi   #angle, degrees     \n",
+    "\n",
+    "print \" The resultant velocity :\",round(V,2),\"km/hour\"\n",
+    "print round(alpha,2),\"°\"\n",
+    "\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {
+    "collapsed": true
+   },
+   "source": [
+    "# example 1.2 Page number 10"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "10.0 KN (to the left)\n",
+      "17.32 KN (downward)\n"
+     ]
+    }
+   ],
+   "source": [
+    "\n",
+    "\n",
+    "\n",
+    "#components of force in horizontal and vertical components. \n",
+    "from math import cos,sin,pi\n",
+    "#variable declaration\n",
+    "\n",
+    "F= 20                        #force in wire, KN\n",
+    "\n",
+    "#calculations\n",
+    "Fx= F*cos(60*pi/180)          \n",
+    "Fy= F*sin(60*pi/180)\n",
+    "\n",
+    "print round(Fx,2),\"KN\" ,\"(to the left)\"\n",
+    "print round(Fy,2), \"KN\" ,\"(downward)\"\n",
+    "\n",
+    "\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {
+    "collapsed": true
+   },
+   "source": [
+    "# example 1.3 Page number 11"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 4,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Component normal to the plane : 9.4 KN\n",
+      "Component parallel to the plane : 3.42 KN\n"
+     ]
+    }
+   ],
+   "source": [
+    "\n",
+    "\n",
+    " #The plane makes an angle of 20° to the horizontal. Hence the normal to the plane makes an angles of 70° to the horizontal i.e., 20° to the vertical\n",
+    "from math import cos,sin,pi\n",
+    "#variable declaration\n",
+    "W= 10                        # black weighing, KN\n",
+    "\n",
+    "#calculations\n",
+    "\n",
+    "Nor= W*cos(20*pi/180)             #Component normal to the plane\n",
+    "para= W*sin(20*pi/180)            #Component parallel to the plane\n",
+    "\n",
+    "print \"Component normal to the plane :\",round(Nor,2),\"KN\"\n",
+    "print \"Component parallel to the plane :\",round(para,2) , \"KN\"\n",
+    "\n",
+    "\n",
+    "\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {
+    "collapsed": true
+   },
+   "source": [
+    "# example 1.4 Page number 11"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 5,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "F1= 100.0 N\n",
+      "F2= 200.0 N\n",
+      "theta= 63.9 °\n"
+     ]
+    }
+   ],
+   "source": [
+    "\n",
+    "\n",
+    "#Let the magnitude of the smaller force be F. Hence the magnitude of the larger force is 2F\n",
+    "\n",
+    "from math import pi,sqrt, acos\n",
+    "#variable declaration\n",
+    "R1=260            #resultant of two forces,N\n",
+    "R2=float(180)          #resultant of two forces if larger force is reversed,N\n",
+    "\n",
+    "\n",
+    "\n",
+    "#calculations\n",
+    "\n",
+    "F=sqrt((pow(R1,2)+pow(R2,2))/10)\n",
+    "F1=F\n",
+    "F2=2*F\n",
+    "theta=acos((pow(R1,2)-pow(F1,2)-pow(F2,2))/(2*F1*F2))*180/pi\n",
+    "\n",
+    "print \"F1=\",F1,\"N\"\n",
+    "print  \"F2=\",F2,\"N\"\n",
+    "print \"theta=\",round(theta,1),\"°\"\n",
+    "\n",
+    "\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {
+    "collapsed": true
+   },
+   "source": [
+    "# example 1.5 Page number 12"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 6,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "F1= 326.35 N\n",
+      "F2= 223.24 N\n"
+     ]
+    }
+   ],
+   "source": [
+    "\n",
+    "\n",
+    "#Let ?ABC be the triangle of forces drawn to some scale\n",
+    "#Two forces F1 and F2 are acting at point A\n",
+    "#angle in degrees '°'\n",
+    "\n",
+    "from math import  sin,pi\n",
+    "  \n",
+    "#variabble declaration\n",
+    "cnv=pi/180\n",
+    "\n",
+    "BAC = 20*cnv                           #Resultant R makes angle with F1    \n",
+    " \n",
+    "ABC = 130*cnv                    \n",
+    "\n",
+    "ACB = 30*cnv   \n",
+    "\n",
+    "R =  500                            #resultant force,N\n",
+    "\n",
+    "#calculations\n",
+    "#sinerule\n",
+    "\n",
+    "F1=R*sin(ACB)/sin(ABC)\n",
+    "F2=R*sin(BAC)/sin(ABC)\n",
+    "\n",
+    "print \"F1=\",round(F1,2),\"N\"\n",
+    "print \"F2=\",round(F2,2),\"N\"\n",
+    "\n",
+    "\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {
+    "collapsed": true
+   },
+   "source": [
+    "# example 1.6 Page number 12"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 7,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "theta= 78.13 °\n",
+      "alpha= 29.29 °\n"
+     ]
+    }
+   ],
+   "source": [
+    "\n",
+    "\n",
+    "#Let ABC  be the triangle of forces,'theta' be the angle between F1 and F2, and 'alpha' be the angle between resultant and F1 \n",
+    "\n",
+    "from math import sin,acos,asin,pi\n",
+    "\n",
+    "#variable declaration\n",
+    "cnv= 180/pi\n",
+    "F1=float(400)                         #all forces are in newtons,'N'\n",
+    "F2=float(260)\n",
+    "R=float(520)\n",
+    "\n",
+    "#calculations\n",
+    "\n",
+    "theta=acos((pow(R,2)-pow(F1,2)-pow(F2,2))/(2*F1*F2))*cnv\n",
+    "\n",
+    "alpha=asin(F2*sin(theta*pi/180)/R)*cnv\n",
+    "\n",
+    "print\"theta=\",round(theta,2),\"°\"\n",
+    "print \"alpha=\",round(alpha,2),\"°\"\n",
+    "\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {
+    "collapsed": true
+   },
+   "source": [
+    "# example 1.7 Page number 13"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 8,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "horizontal component= 2814.2 N\n",
+      "Vertical component =  1039.2 N\n",
+      "Component along crank = 507.1 N\n",
+      "Component normal to crank= 2956.8 N\n"
+     ]
+    }
+   ],
+   "source": [
+    "\n",
+    "\n",
+    "#The force of 3000 N acts along line AB. Let AB make angle alpha with horizontal.\n",
+    "\n",
+    "from math import cos,sin,pi,asin,acos\n",
+    "\n",
+    "#variable declaration\n",
+    "F=3000                        #force in newtons,'N'\n",
+    "BC=80                         #length of crank BC, 'mm'\n",
+    "AB=200                        #length of connecting rod AB ,'mm'\n",
+    "theta=60*pi/180               #angle b/w BC & AC\n",
+    "\n",
+    "#calculations\n",
+    "\n",
+    "alpha=asin(BC*sin(theta)/200)*180/pi\n",
+    "\n",
+    "HC=F*cos(alpha*pi/180)                    #Horizontal component \n",
+    "VC= F*sin(alpha*pi/180)                   #Vertical component \n",
+    "\n",
+    "#Components along and normal to crank\n",
+    "#The force makes angle alpha + 60  with crank.\n",
+    "alpha2=alpha+60\n",
+    "CAC=F*cos(alpha2*pi/180)             # Component along crank \n",
+    "CNC= F*sin(alpha2*pi/180)             #Component normal to crank \n",
+    "\n",
+    "\n",
+    "print \"horizontal component=\",round(HC,1),\"N\"\n",
+    "print \"Vertical component = \",round(VC,1),\"N\"\n",
+    "print \"Component along crank =\",round(CAC,1),\"N\"\n",
+    "print \"Component normal to crank=\",round(CNC,1),\"N\""
+   ]
+  }
+ ],
+ "metadata": {
+  "anaconda-cloud": {},
+  "kernelspec": {
+   "display_name": "Python [Root]",
+   "language": "python",
+   "name": "Python [Root]"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 2
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython2",
+   "version": "2.7.12"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/sample_notebooks/KARTHIKEYAN S/CHAPTER_1.ipynb b/sample_notebooks/KARTHIKEYAN S/CHAPTER_1.ipynb
new file mode 100644
index 00000000..8f5d99cb
--- /dev/null
+++ b/sample_notebooks/KARTHIKEYAN S/CHAPTER_1.ipynb
@@ -0,0 +1,104 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# CHAPTER 1:Fundmentals"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## EXAMPLE 1.1,PAGE NUMBER:3"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 6,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Carnot COP= 6.0 (error)\n"
+     ]
+    }
+   ],
+   "source": [
+    "\n",
+    "import math\n",
+    "\n",
+    "#Variable Declaration\n",
+    "T_0=-5+273;\n",
+    "T_1=35+273;\n",
+    "\n",
+    "#Calculation\n",
+    "COP=(T_0)/(T_1-T_0);# Coefficient of performance\n",
+    "print \"Carnot COP=\",round(COP,2),\"(error)\"\n",
+    "\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## EXAMPLE 1.2,PAGE NUMBER:4"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 5,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "The average specific heat capacity is 4.186 kJ/(kg K)\n"
+     ]
+    }
+   ],
+   "source": [
+    "\n",
+    "import math\n",
+    "\n",
+    "#Variable Declaration\n",
+    "T_f=80;# Final Temperature in °C\n",
+    "T_i=0;# Initial Temperature in °C\n",
+    "h_f=334.91;#The specific enthalpy of water in kJ/kg\n",
+    "\n",
+    "#Calculation\n",
+    "C=h_f/(T_f-T_i);# The average specifi c heat capacity in kJ/(kg K)\n",
+    "print \"The average specific heat capacity is\",round(C,3),\"kJ/(kg K)\"\n"
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python 2",
+   "language": "python",
+   "name": "python2"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 2
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython2",
+   "version": "2.7.11"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/sample_notebooks/NityaL/Sample-Chapter_26.ipynb b/sample_notebooks/NityaL/Sample-Chapter_26.ipynb
new file mode 100644
index 00000000..4ae1c760
--- /dev/null
+++ b/sample_notebooks/NityaL/Sample-Chapter_26.ipynb
@@ -0,0 +1,226 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# Chapter 26:CHARGE AND MATTER"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {
+    "collapsed": true
+   },
+   "source": [
+    "# Example 26.1 Magnitude of total charges in a copper penny"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 5,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      " Magnitude of the charges in coulombs is  133687.50000000003\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Example 1.1\n",
+    "\n",
+    "m =3.1 #mass of copper penny in grams\n",
+    "e =4.6*10** -18 #charge in coulombs\n",
+    "N0 =6*10**23 #avogadro’s number atoms / mole\n",
+    "M =64  #molecular weight of copper in gm/ mole\n",
+    "\n",
+    "#Calculation\n",
+    "N =( N0 * m ) / M  #No. of copper atoms in penny\n",
+    "q = N * e  # magnitude of the charges in coulombs\n",
+    "print (\" Magnitude of the charges in coulomb is \",q )"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {
+    "collapsed": true
+   },
+   "source": [
+    "# Example 26.2 Separation between total positive and negative charges"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      " Separation between total positive and negative charges in meters is  5813776741.499454\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Example 2\n",
+    "\n",
+    "import math\n",
+    "\n",
+    "F =4.5 #Force of attraction in nt\n",
+    "q =1.3*10**5 #total charge in coulomb\n",
+    "r = q * math.sqrt ((9*10**9) / F ) ;\n",
+    "print(\" Separation between total positive and negative charges in meters is \",r )"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {
+    "collapsed": true
+   },
+   "source": [
+    "# Example 26.3 Force acting on charge q1"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 8,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "X component of resultant force acting on q1 in nt is 2.0999999999999996\n",
+      "Y component of resultant force acting on q1 in nt is -1.5588457268119893\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Example 3\n",
+    "\n",
+    "import math\n",
+    "\n",
+    "#given three charges q1,q2,q3\n",
+    "q1=-1.0*10**-6 #charge in coul\n",
+    "q2=+3.0*10**-6 #charge in coul\n",
+    "q3=-2.0*10**-6 #charge in coul\n",
+    "r12=15*10**-2 #separation between q1 and q2 in m\n",
+    "r13=10*10**-2 # separation between q1 and q3 in m\n",
+    "angle=math.pi/6 #in degrees\n",
+    "F12=(9.0*10**9)*q1*q2/(r12**2) #in nt\n",
+    "F13=(9.0*10**9)*q1*q3/(r13**2) #in nt\n",
+    "F12x=-F12  #ignoring signs of charges\n",
+    "F13x=F13*math.sin(angle);\n",
+    "F1x=F12x+F13x\n",
+    "F12y=0 #from fig.263\n",
+    "F13y=-F13*math.cos(angle);\n",
+    "F1y=F12y+F13y #in nt\n",
+    "print(\"X component of resultant force acting on q1 in nt is\",F1x)\n",
+    "print(\"Y component of resultant force acting on q1 in nt is\",F1y)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {
+    "collapsed": true
+   },
+   "source": [
+    "# Example 26.4 Electrical and Gravitational force between two particles"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 9,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Coulomb force in nt is 8.202207191171238e-08\n",
+      "Gravitational force in nt is 3.689889640441438e-47\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Example 4\n",
+    "\n",
+    "r=5.3*10**-11 #distance between electron and proton in the hydrogen atom in meter\n",
+    "e=1.6*10**-19 #charge in coul\n",
+    "G=6.7*10**-11 #gravitatinal constant in nt-m2/kg2\n",
+    "m1=9.1*10**-31 #mass of electron in kg\n",
+    "m2=1.7*10**-27 #mass of proton in kg\n",
+    "F1=(9*10**9)*e*e/(r**2) #coulomb's law\n",
+    "F2=G*m1*m2/(r**2) #gravitational force\n",
+    "print(\"Coulomb force in nt is\",F1)\n",
+    "print(\"Gravitational force in nt is\",F2)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {
+    "collapsed": true
+   },
+   "source": [
+    "# Example 26.5 Repulsive force between two protons in a nucleus of iron"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Repulsive coulomb force F  14.4 nt\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Example 5\n",
+    "\n",
+    "r=4*10**-15 #separation between proton annd nucleus in iron in meters\n",
+    "q=1.6*10**-19 #charge in coul\n",
+    "F=(9*10**9)*(q**2)/(r**2) #coulomb's law\n",
+    "print(\"Repulsive coulomb force F \",F,'nt')"
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python [Root]",
+   "language": "python",
+   "name": "Python [Root]"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 2
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython2",
+   "version": "2.7.12"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}