# Chapter 5 Properties of matter¶

## Example 5.1 Page no 76¶

In [1]:
#given
m=1                                  #Mass of torsional pendulum in kg
R=0.06                               #Radius of torsional pendulum in m
l=1.2                                #Length of the wire in m
r=0.0008                             #Radius of wire in m
S=(9*10**9)                          #Modulus of rigidity of the material in N/m^2

#Calculations
import math
I=(1/2.0)*m*R**2
C=(3.14*S*r**4)/(2*l)
T=2*3.14*math.sqrt(I/C)

#Output
print"Period of pendulum is ",round(T,1),"s"

Period of pendulum is  3.8 s


## Example 5.2 Page no 76¶

In [4]:
#given
l=0.8                                          #Length of the wire in m
d=(1.8*10**-3)                                 #Diameter of the wire in m
a=1.5                                          #Angle of twist in degrees
S=(1.8*10**11)                                 #Modulus of rigidity of the material in N/m^2

#Calculations
r=(a*3.14)/180.0
W=((3.14*S*(d/2.0)**4*r**2)/(4*l))/10.0**-5

#Output
print"Work required to twist the wire is ",round(W,2),"*10^-5 J"

Work required to twist the wire is  7.93 *10^-5 J


## Example 5.3 Page no 76¶

In [7]:
#given
l=2                                                   #Length of wire in m
d=(0.4*10**-3)                                        #Diameter of the wire in m
x=(1.03*10**-3)                                       #Extension in length in m
L=2                                                   #Load in kg
C=(4.52*10**-6)                                       #Couple in N/m
a=0.03                                                #Twist angle in radians

#Calculations
Y=((L*9.8*l)/(x*3.14*(d/2.0)**2))/10**11
S=((C*2*l)/(3.14*(d/2.0)**4*a))/10**11
s=(Y/(2*S))-1

#Output
print"Youngs modulus is ",round(Y,2),"*10**11 N/m^2"
print"Modulus of rigidity is ",round(S,2),"*10**11 N/m^2"
print"Poissons ratio is ",round(s,2)

Youngs modulus is  3.03 *10**11 N/m^2
Modulus of rigidity is  1.2 *10**11 N/m^2
Poissons ratio is  0.26


## Example 5.4 Page no 76¶

In [8]:
#given
r=0.003                                            #Radius of drop of glycerine in m
T=(63.1*10**-3)                                    #Surface tension of glycerine in N/m

#Calculations
P=((2*T)/r)

#Output
print"Excess pressure inside the drop of glycerine is ",round(P,2),"N/m^2"

Excess pressure inside the drop of glycerine is  42.07 N/m^2


## Example 5.5 Page no 76¶

In [5]:
#given
r1=0.001                                            #Initial radius in m
r2=0.004                                            #Final radius in m
t=2*10**-3                                          #Time in s
s=(7*10**-2)                                        #Surface tension of water in N/m

#Calculations
P=((2*s)*((1/r2)-(1/r1)))/(t*10**4)

#Output
print"Rate of change of pressure is ",P,"*10**4 N/m**2 s"

Rate of change of pressure is  -5.25 *10**4 N/m**2 s


## Example 5.6 Page no 77¶

In [9]:
#given
d=0.02                                            #Diamter of soap bubble in m
s=(25*10**-3)                                     #Surface tension in N/m
#Initial surface area of the bubble is zero and final area is 2*4*pie*r^2 where r is the radius of the bubble

#Calculations
W=(s*2*4*3.14*(d/2.0)**2)/10.0**-5

#Output
print"Work done in blowing a soap bubble is ",W,"*10**-5 J"

Work done in blowing a soap bubble is  6.28 *10**-5 J


## Example 5.7 Page no 77¶

In [8]:
#given
r=0.01                                                 #Radius of liquid drop in m
n=500                                                  #Number of drops
s=(63*10**-3)                                          #Surface tension in N/m

#Calculations
r1=(((4*3.14*r**3)/3.0)/((n*4*3.14)/3.0))**(1/3.0)
As=(n*4*3.14*r1**2)
A=4*3.14*r**2
W=(s*(As-A))/10.0**-4

#Output
print"Energy required to break up a drop of a liquid is ",round(W,1),"*10**-4 J"

Energy required to break up a drop of a liquid is  5.5 *10**-4 J


## Example 5.8 Page no 77¶

In [10]:
#given
d=0.04                                        #Inside diameter of garden hose in m
D=0.01                                        #Diamter of nozzle opening in m
v1=0.6                                        #speed of flow of water in the hose in m/s

#calculations
a=3.14*(d/2.0)**2
A=3.14*(D/2.0)**2
v2=(v1*a)/A

#Output
print"Speed of flow through the nozzle is ",v2,"m/s"

Speed of flow through the nozzle is  9.6 m/s