Physics Times 07.2019

2.Sol: Young’s modulus

W l

Y=

AΔl



l, A and l are the same for both the wires.

Y W

2

1

s s

b b

Y W

=

Y W



(^) W Ws: 2:1b
3.Sol:
1
B
Compressibility

1 h g
compressibility V
V


 
  
 
V
h g Compressibility
V
 
  
 
3 3 11
V
2.7 10 10 10 45.4 10
V
       
(^)  1.23 10^2
4.Sol: Young’s modulus is given by
F
Y
A
l
l



( Hooke’s Law) (i)
As V A  l constant (ii)
From Eqs. (i) and (ii), we get
F^2
Y
V
l
l



2
F
V Y
  l l

(^)  l l^2
5.Sol:As
F L mg.L
Y
 L A L A

 
 
or
mgL
L
YA
 
L
L ,
A
  which is maximum for
option (b)
FLUID MECHANICS
1.Sol: The cylinder present in two liquids are
shown in the figure.
4.Sol:
As the cylinder is in equilibrium F =F +Fg b 1 b 2
LAdg pL A n g 1 p LA g         
       d 1 p pn 1 n 1 p     
2.Sol: From equation of continuity
2
2 2
(^112)
R
R n r
nr
v
    v v v
3.Sol: Applying Bernoulli’s theorem just above
and just below the roof,
(^20)
1
P P 0
2
   v   0 2
1
P P P
2
   v
Hence lift of the roof
F P.A^1 A^2
2
   v
SURFACE TENSION
2.Sol: Let R is the radius of the bigger drop and
N drops are combined to form the bigger drop.
(^4343)
3 3
N r  R
Nr R^3 ^3 (i)
The energy released
E T A A   i f
E T N r    4 2 4 R^2 
2 2
4 2 1
Nr
R T
R
 
   
 
3 2
3
4 1
3
3
Nr
R T
R R
   
    
   
3 VT 1 1
r R
  
  
2 2 2 2 2
W       8 (S R R 2 1 ) 8 [(2 )S R R] 24 R S
5.Sol: Mass of liquid in capillary tube
2.Sol:
3.Sol:
4.Sol:
5.Sol:
FLUID MECHANICS
1.Sol:
2.Sol:
3.Sol:
SURFACE TENSION
2.Sol:
4.Sol:
5.Sol:
M R h^2 R^2 2 cosT
Rg

  

 M R. If radius becomes double, then
mass will become twice.