Mechanical Properties of Fluids Practice Questions - DPP for JEE

 Page 1


1. (a) Inflow rate of volume of the liquid = Outflow rate of volume of the
liquid
pR
2
V = npr
2
(v) ? v = 
2. (a) When the bubble gets detached,
Buoyant force = force due to surface tension
Force due to excess pressure =  upthrust
Access pressure in air bubble = 
? ? 
3. (d) From the figure it is clear that liquid 1 floats on liquid  2.  The
lighter liquid floats over heavier liquid. Therefore we can conclude
that ?
1
 < ?
2
Also ?
3
 < ?
2
 otherwise the ball would have sink to the bottom of the jar.
Also ?
3
 > ?
1
 otherwise the ball would have floated in liquid 1. From the
above discussion we conclude that
?
1
 < ?
3
 < ?
2
.
4. (c) Upthrust = weight of 40 cm
3
 of water
= 40 g = down thrust on water
5. (d) T
1
 + T cos (p – ?) = T
2
Page 2


1. (a) Inflow rate of volume of the liquid = Outflow rate of volume of the
liquid
pR
2
V = npr
2
(v) ? v = 
2. (a) When the bubble gets detached,
Buoyant force = force due to surface tension
Force due to excess pressure =  upthrust
Access pressure in air bubble = 
? ? 
3. (d) From the figure it is clear that liquid 1 floats on liquid  2.  The
lighter liquid floats over heavier liquid. Therefore we can conclude
that ?
1
 < ?
2
Also ?
3
 < ?
2
 otherwise the ball would have sink to the bottom of the jar.
Also ?
3
 > ?
1
 otherwise the ball would have floated in liquid 1. From the
above discussion we conclude that
?
1
 < ?
3
 < ?
2
.
4. (c) Upthrust = weight of 40 cm
3
 of water
= 40 g = down thrust on water
5. (d) T
1
 + T cos (p – ?) = T
2
? cos (p – ?) = 
?
?
6. (a) The condition for terminal speed (v
t
) is
Weight = Buoyant force + Viscous force
7. (a) As A
1
v
1
 = A
2
v
2
 (Principle of continuity)
or,  (Efflux velocity = )
?     or
8. (b) Pressure inside tube = 
?  (since r
2
 > r
1
)
Hence pressure on side 1 will be greater than side 2. So air from end 1
flows towards end 2
Page 3


1. (a) Inflow rate of volume of the liquid = Outflow rate of volume of the
liquid
pR
2
V = npr
2
(v) ? v = 
2. (a) When the bubble gets detached,
Buoyant force = force due to surface tension
Force due to excess pressure =  upthrust
Access pressure in air bubble = 
? ? 
3. (d) From the figure it is clear that liquid 1 floats on liquid  2.  The
lighter liquid floats over heavier liquid. Therefore we can conclude
that ?
1
 < ?
2
Also ?
3
 < ?
2
 otherwise the ball would have sink to the bottom of the jar.
Also ?
3
 > ?
1
 otherwise the ball would have floated in liquid 1. From the
above discussion we conclude that
?
1
 < ?
3
 < ?
2
.
4. (c) Upthrust = weight of 40 cm
3
 of water
= 40 g = down thrust on water
5. (d) T
1
 + T cos (p – ?) = T
2
? cos (p – ?) = 
?
?
6. (a) The condition for terminal speed (v
t
) is
Weight = Buoyant force + Viscous force
7. (a) As A
1
v
1
 = A
2
v
2
 (Principle of continuity)
or,  (Efflux velocity = )
?     or
8. (b) Pressure inside tube = 
?  (since r
2
 > r
1
)
Hence pressure on side 1 will be greater than side 2. So air from end 1
flows towards end 2
9. (c) Velocity of ball when it strikes the water surface
...(i)
Terminal velocity of ball inside the water
...(ii)
Equation (i) and (ii) we get 
10. (a) Fluid resistance is given by R = 
When two capillary tubes of same size are joined in parallel, then
equivalent fluid resistance is
=  = 
Rate of flow  =  = X   
11. (c) Vertical distance covered by water before striking ground = (H –
h). Time taken is,  : Horizontal velocity of water
coming out of hole at P, 
?  Horizontal range = 
= 
12. (a) Because film tries to cover minimum surface area.
Page 4


1. (a) Inflow rate of volume of the liquid = Outflow rate of volume of the
liquid
pR
2
V = npr
2
(v) ? v = 
2. (a) When the bubble gets detached,
Buoyant force = force due to surface tension
Force due to excess pressure =  upthrust
Access pressure in air bubble = 
? ? 
3. (d) From the figure it is clear that liquid 1 floats on liquid  2.  The
lighter liquid floats over heavier liquid. Therefore we can conclude
that ?
1
 < ?
2
Also ?
3
 < ?
2
 otherwise the ball would have sink to the bottom of the jar.
Also ?
3
 > ?
1
 otherwise the ball would have floated in liquid 1. From the
above discussion we conclude that
?
1
 < ?
3
 < ?
2
.
4. (c) Upthrust = weight of 40 cm
3
 of water
= 40 g = down thrust on water
5. (d) T
1
 + T cos (p – ?) = T
2
? cos (p – ?) = 
?
?
6. (a) The condition for terminal speed (v
t
) is
Weight = Buoyant force + Viscous force
7. (a) As A
1
v
1
 = A
2
v
2
 (Principle of continuity)
or,  (Efflux velocity = )
?     or
8. (b) Pressure inside tube = 
?  (since r
2
 > r
1
)
Hence pressure on side 1 will be greater than side 2. So air from end 1
flows towards end 2
9. (c) Velocity of ball when it strikes the water surface
...(i)
Terminal velocity of ball inside the water
...(ii)
Equation (i) and (ii) we get 
10. (a) Fluid resistance is given by R = 
When two capillary tubes of same size are joined in parallel, then
equivalent fluid resistance is
=  = 
Rate of flow  =  = X   
11. (c) Vertical distance covered by water before striking ground = (H –
h). Time taken is,  : Horizontal velocity of water
coming out of hole at P, 
?  Horizontal range = 
= 
12. (a) Because film tries to cover minimum surface area.
13. (c) Pressure at interface A must be same from both the sides to be in
equilibrium.
? (R cos a + R sin a)d
2
g = (R cos a – R sin a)d
1
g
?
14. (c) Angle of contact ?
when water is on a waxy or oily surface
T
SA
 < T
SL
 cos ? is negative i.e.,  90° < ? < 180°
i.e., angle of contact ? increases
And for ? > 90º liquid level in capillary tube fall. i.e., h decreases
15. (c) Sum of volumes of 2 smaller drops
= Volume of the bigger drop
   
Surface energy = T.4pR
2
= T4p2
2/3
r
2
 = T.2
8/3
 pr
2
16. (a) When a body falls through a viscous liquid, its velocity increases
due to gravity but after some time its velocity becomes uniform
because of viscous force becoming equal to the gravitational force.
Viscous force itself is a variable force which increases as velocity
increases, so curve (a) represents the correct alternative.
17. (c)
Page 5


1. (a) Inflow rate of volume of the liquid = Outflow rate of volume of the
liquid
pR
2
V = npr
2
(v) ? v = 
2. (a) When the bubble gets detached,
Buoyant force = force due to surface tension
Force due to excess pressure =  upthrust
Access pressure in air bubble = 
? ? 
3. (d) From the figure it is clear that liquid 1 floats on liquid  2.  The
lighter liquid floats over heavier liquid. Therefore we can conclude
that ?
1
 < ?
2
Also ?
3
 < ?
2
 otherwise the ball would have sink to the bottom of the jar.
Also ?
3
 > ?
1
 otherwise the ball would have floated in liquid 1. From the
above discussion we conclude that
?
1
 < ?
3
 < ?
2
.
4. (c) Upthrust = weight of 40 cm
3
 of water
= 40 g = down thrust on water
5. (d) T
1
 + T cos (p – ?) = T
2
? cos (p – ?) = 
?
?
6. (a) The condition for terminal speed (v
t
) is
Weight = Buoyant force + Viscous force
7. (a) As A
1
v
1
 = A
2
v
2
 (Principle of continuity)
or,  (Efflux velocity = )
?     or
8. (b) Pressure inside tube = 
?  (since r
2
 > r
1
)
Hence pressure on side 1 will be greater than side 2. So air from end 1
flows towards end 2
9. (c) Velocity of ball when it strikes the water surface
...(i)
Terminal velocity of ball inside the water
...(ii)
Equation (i) and (ii) we get 
10. (a) Fluid resistance is given by R = 
When two capillary tubes of same size are joined in parallel, then
equivalent fluid resistance is
=  = 
Rate of flow  =  = X   
11. (c) Vertical distance covered by water before striking ground = (H –
h). Time taken is,  : Horizontal velocity of water
coming out of hole at P, 
?  Horizontal range = 
= 
12. (a) Because film tries to cover minimum surface area.
13. (c) Pressure at interface A must be same from both the sides to be in
equilibrium.
? (R cos a + R sin a)d
2
g = (R cos a – R sin a)d
1
g
?
14. (c) Angle of contact ?
when water is on a waxy or oily surface
T
SA
 < T
SL
 cos ? is negative i.e.,  90° < ? < 180°
i.e., angle of contact ? increases
And for ? > 90º liquid level in capillary tube fall. i.e., h decreases
15. (c) Sum of volumes of 2 smaller drops
= Volume of the bigger drop
   
Surface energy = T.4pR
2
= T4p2
2/3
r
2
 = T.2
8/3
 pr
2
16. (a) When a body falls through a viscous liquid, its velocity increases
due to gravity but after some time its velocity becomes uniform
because of viscous force becoming equal to the gravitational force.
Viscous force itself is a variable force which increases as velocity
increases, so curve (a) represents the correct alternative.
17. (c)
New mass  m
2
 = 
= 
18. (d) The excess pressure inside a bubble of soap is given by . It
means excess pressure inside the bubbles A and C is more than
inside B. So, air will go towards B from A and C. So, A and C will
start collapsing with the increasing volume of B.
19. (c)
20. (a)
21. (8) Inside pressure must be  greater than outside pressure in bubble.
This excess pressure is provided by charge on bubble.
;    
22. (0.1) Terminal velocity, 
23. (10.67) Let V
i 
be the volume of the iceberg inside sea water and  V is
the total volume of iceberg.