Class 11-science H C VERMA Solutions Physics Chapter 14 - Some Mechanical Properties of Matter
Some Mechanical Properties of Matter Exercise 300
Solution 1
m=10kg ;
l=3m ;
A=4mm2;
Y=2×1011 N/m2
Solution :
a) Stress=
b) Y=
c) Strain=
Solution 2
m=100kg;
l=2m;
=
Solution:
a) 
b) Strain=
c) Compression
Solution 3
Solution 4
Lst=Lcu
Ast=Acu
Fst=Fcu
a) Stress=
So, stress in both wire is same.
b) strain
Solution 5
Lst=Lcu
Ast=Acu
Fst=Fcu=mg
Strain=
Solution 6
(a) Let mass put in hanger is m
For lower wire
TL=(m1+m)g;
;
AL=0.003cm2=3
For upper wire
TU=( m2 + m1+m)g;
;
Au=0.006cm2=
So, lower wire will break first. So maximum mass put in hanger is 14kg.
(b) If m1=10kg and m2=36kg
For lower wire,
For upper wire,
So, upper wire will break first. So, maximum mass put in hanger is 2kg.
Solution 7
F=100N;
L=2m;
A=2cm2
Solution 8
Y=
L=2m;
A=4cm2
Solution 9
Translatory motion equations
m2g-T=m2a 1)
T-
= m1a
2)
Adding both equation(1) and (2)
=( m1 +m2)a
Put in equation (1) and (2)
Now,
Strain=
Strain=
Solution 10
At equilibrium T=mg
Now, when it is moved by angle θ and released, because of centrifugal force at lowest point, Tension becomes-
T'=mg+
So, change in tension
By energy Conservation
0+
mgl(1-cosθ)=
Now, 
L=4m
m=20kg;
Y=2
Some Mechanical Properties of Matter Exercise 301
Solution 11
Let depression is wire be x
Now, length of wire becomes=2
Change in length=2
By binomial expansion
From equilibrium of forces,
Now,
On solving,
Solution 12
A=0.01
T=20N
Y=1.1×
e=0.32
Now,
Again,
Solution 13
Solution 14
Change in
density=
=1032-1030
=2
Solution 15
Solution 16
F
Solution 17
a) 
b) 
c) 
Solution 18
a) Force by air above it
b) By mercury below it
c) By mercury surface in contact with it
Solution 19
a) 
b) 
c) 
Solution 20
So,
Solution 21
Not rise in bare meter tube=76-0.5=75.5cm
Solution 22
Pressure in tube 5cm below water surface is
So, excess Pressure
Solution 23
Surface energy=TA
Solution 24
Radius of big drop
Let radius of
small drop be
Here,
Volume of big drop = Volume of small drop
Increase in
surface energy=
Solution 25
a) 
b)Initially
…………………(1)
Now,
…………..….(2)
Dividing Both equation
Solution 26
a)
b) 
Solution 27
Force due to surface Tension = Force by gravity
Solution 28
After melting the Ice cube will acquire shape of sphere as it is kept in gravity free space and due to surface tension.
Volume of cube= volume of sphere
Surface
Area of spherical drop=
Solution 29
Considering a small section on thread subtending an angle of 2dθ at center.
Force due to soap solution on this section=2T(2Rdθ)
This
force is balanced by tension in thread= 2T'sin d
=2T'dθ
By equilibrium of force
2T(2Rdθ)= 2T'dθ
T'=2TR
T'=2T(
)
T'=
T'=
Solution 30
a) Viscous Force
ηrv
b)
hydrostatic force=
c) At the time of terminal velocity
Download force= Upward force
ηrv
Solution 31
Some Mechanical Properties of Matter Exercise 302
Solution 32
V=
m/s
R=1
m
poise=10-3poisullie
Reynold's number
Reynold number is less than 1200. Therefore, it is a steady Flow.
