Fluids properties

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Chapter 10
Mechanical Properties of Fluids
SECTION - A
Objective Type Questions
1. The term ‘fluid’ is used for
(1) Liquids only (2) Gases only
(3) A mixture of liquid and gas only (4) Both liquids and gases
Sol. Answer (4)
Substances that can flow is called fluid. Thus both liquids and gases are fluids.
2. Select wrong statement about pressure
(1) Pressure is a scalar quantity (2) Pressure is always compressive in nature
(3) Pressure at a point is same in all directions (4) None of these
Sol. Answer (4)
Pressure is scalar as it is not added vectorially. Pressure is compressive in nature, it is same in all the
directions at a point.
3. Gauge pressure
(1) May be positive (2) May be negative (3) May be zero (4) All of these
Sol. Answer (4)
Gauge pressure depends on the reference chosen, it can be positive, negative or zero.
4. Figure shows two containers P and Q with same base area A and each filled upto same height with same
liquid. Select the correct alternative
A
px
( )P
A
py
( )Q
(1) px = py (2) px > py (3) py > px (4) Cannot say
Solutions

98 Mechanical Properties of Fluids Solution of Assignment (Set-2)
Sol. Answer (1)
The level of water above both points is same so by hydrostatic paradox.
px = py
5. The pressure of confined air is p. If the atmospheric pressure is P, then
(1) P is equal to p
(2) P is less than p
(3) P is greater than p
(4) P may be less or greater than p depending on the mass of the confined air
Sol. Answer (2)
Pressure at point A = P (Hydrostatic paradox)
and P + wgh = pressure at A = p
 P = p – wgh
P
A ph p
air
WaterSo, P < p
6. Figure shows a container filled with a liquid of density . Four points A, B, C and D lie on the diametrically opposite
points of a circle as shown. Points A and C lie on vertical line and points B and D lie on horizontal line. The
incorrect statement is (pA, pB, pC, pD are absolute pressure at the respective points)
A
B
C
D
(1) pD = pB (2) pA < pB = pD < pC (3)
2
AC
BD
pp
pp

 (4)
2
C A
D B
p p
p p

 
Sol. Answer (3)
Points at same height have same pressure, points with height difference say 'h' will have difference of gh.
Let radius of circle is r
pA = p0 + hg
pB = pD = p0 + (h + r)g
pC = p0 + (h + 2r)g
Then,
A
D
C
B
h
p0
r
r
pC + pA = [p0 + (h + 2r)g] – [p0 + hg]
= p0 + hg + 2rg + p0 + hg
= 2[p0 + (h + r)g]
0 ( )
2
C Ap p
p h r g

   
2
C A
B D
p p
p p

 
i.e., option (1), (2) and (4) gives correct statement but incorrect statement is (3)

99Solution of Assignment (Set-2) Mechanical Properties of Fluids
7. The volume of an air bubble is doubled as it rises from the bottom of lake to its surface. The atmospheric pressure
is 75 cm of mercury. The ratio of density of mercury to that of lake water is
3
40
. The depth of the lake in metre is
(1) 10 (2) 15 (3) 20 (4) 25
Sol. Answer (1)
2P0 = P0 + gh
 P0 = gh
 P0 = 75 cm mercury [Atmospheric pressure]
 mercury water
75
100
g g h      

75
100
m
w
h

 


40 75 40
(given)
3 100 3
m
w
h
⎡ ⎤
  ⎢ ⎥
⎣ ⎦
∵
 h = 10 m
8. A beaker containing a liquid of density  moves up with an acceleration ‘a’. The pressure due to the liquid at a
depth h below free surface of the liquid is
(1) hg (2) h (g – a)
(3) h (g + a) (4) 2
g a
h g
g a
⎛ ⎞
 ⎜ ⎟
⎝ ⎠
Sol. Answer (3)
Due to upward acceleration pseudo force will act downwards so value of acceleration due to gravity will increase
by 'a'
 g' = (g + a)
P = g'h
 P =  (g + a)h (Substitute g')
9. A barometer kept in an elevator reads 76 cm when it is at rest. If the elevator goes up with some acceleration, the
reading will be
(1) 76 cm (2) > 76 cm (3) < 76 cm (4) Zero
Sol. Answer (3)
When elevator goes up with some acceleration upward, due to pseudo force acting downwards. Value of g
increases to g'.
If g increases to g', i.e., g' = g + a
∵ P = gh = (9 + a)h' = constant
Then, h' < h
i.e., h' < 76 cm

10. In a hydraulic jack as shown, mass of the car W = 800 kg, A1 = 10 cm2, A2 = 10 m2. The minimum force F required
to lift the car is
W
A2
A1
F
(1) 1 N (2) 0.8 N (3) 8 N (4) 16 N
Sol. Answer (2)
Pressure in a liquid is divided equally so we can say pressure at both the pistons should be same
 1 2
1 2
F F
A A

Substituting values,
1
2
1
2 4 2
2
2
Where,
10 cm
10 m 10 10 cm
8000 N
F F
A
A
F
⎧
⎪
⎪
⎪
⎨
⎪
  ⎪
⎪ ⎩
4
8000
10 10 10
F


 F = 0.8 N 2
Take
10 m/sg
⎧ ⎫⎪ ⎪
⎨ ⎬
⎪ ⎪⎩ ⎭
11. A wooden cube just floats inside water with a 200 gm mass placed on it. When the mass is removed, the cube
floats with its top surface 2 cm above the water level. What is the side of the cube ?
(1) 6 cm (2) 8 cm (3) 10 cm (4) 12 cm
Sol. Answer (3)
Mass × g = Volume of part of cube ×  × g
 200 × g = L2 (2 × w × g)
 100 = L2 {∵w = 1}
 10 cm = L
200 gm
L
L
( –2) cmL
L
2 cm
From the two figures we can see that the 200 gm block
is provided with required buoyant force but a part of
cube which is afloat in 2nd figure.
12. A block of steel of size 5 × 5 × 5 cm3 is weighed in water. If relative density of steel is 7, its apparent weight is
(1) 6 × 5 × 5 × 5 g wt (2) 4 × 4 × 4 × 7 g wt
(3) 5 × 5 × 5 × 7 g wt (4) 4 × 4 × 4 × 6 g wt
Sol. Answer (1)
Apparent weight = s vb g – w vb g
Where,
density of steel
density of water
volume of block
(side side side)
s
w
bv
⎧
⎪ ⎪
⎨
 ⎪
⎪ ⎩
 
= vbg (s – w)
= 5 × 5 × 5 × g × (7 – 1)
1
7 (given)
w
s
 ⎡
⎢
 ⎣
= 6 × 5 × 5 × 5 × g wt

13. A block of wood floats in water with th
5
4
of its volume submerged, but it just floats in another liquid. The density
of liquid is (in kg/m3)
(1) 750 (2) 800 (3) 1000 (4) 1250
Sol. Answer (2)
4
5
vb × w × g = vb × b × g
3
Where,
volume of block
density of water 1000 kg/m
density of block
b
w
b
v
⎧
⎪ ⎪
⎨
  ⎪
⎪ ⎩

5
4
w
b



34
1000 800 kg/m
5
b   
And when block is put in liquid of density l it just floats
So, vb × b × g = vb × l × g
 b = l
So, l = 800 kg/m3
14. A cubical block is floating in a liquid with one fourth of its volume immersed in the liquid. If whole of the system
accelerates upward with acceleration g/4, the fraction of volume immersed in the liquid will be
(1) 1/4 (2) 1/2 (3) 3/4 (4) 2/3
Sol. Answer (1)
Upward acceleration just causes the acceleration due to gravity increases by some value, but since the term of 'g'
gets cancelled out in the buoyancy equation.
Volume immersed × w × g = Total volume × cube × g
So, increasing it will not have any effect on the immersed volume.
15. A body of density  is dropped from rest from a height h into a lake of density , where  > . Neglecting all
dissipative forces, the maximum depth to which the body sinks before returning to float on surface
(1)

h
(2)

h
(3)

h
(4)

h
Sol. Answer (3)
Between point 2 and 3
2
0 0
1
2
P gh v gh P        [P0 = atmospheric pressure]
 gh' + gh = gh' 2v gh⎡ ⎤⎣ ⎦∵
1 P
h
2
3

h
2v gh
So,
h
h

 
  

16. Aboat carrying a number of stones is floating in a water tank. If the stones are unloaded into water, the water level
in the tank will
(1) Remain unchanged
(2) Rise
(3) Fall
(4) Rise or fall depends on the number of stones unloaded
Sol. Answer (3)
Previously when stones are on the boat they are increasing the weight on the boat and to balance this weight boat
needs to generate buoyancy force by displacing more water, but when stones are removed the boat starts displacing
less amount of water hence the level of water in tank falls.
Decrease in level
17. Ablock of ice is floating in a liquid of specific gravity 1.2 contained in a beaker. When the ice melts completely, the
level of liquid in the vessel
(1) Increases (2) Decreases
(3) Remain unchanged (4) First increases then decreases
Sol. Answer (1)
Density of ice is less than water and density of liquid is more than water. So even when ice melts the level will rise.
If liquid > water then level (liquid + water) will rise.
18. Two liquids having densities d1 and d2 are mixed in such a way that both have same mass. The density of the
mixture is
(1)
1 2
2
d d
(2)
1 2
1 2
d d
d d

(3)
1 2
1 2
d d
d d (4)
1 2
1 2
2d d
d d
Sol. Answer (4)
Let each have mass = M and densities d1 and d2
mix
mix
mix
1 2
M M M
d
V M M
d d

 
⎛ ⎞ ⎛ ⎞
⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
1 2
1 2
2d d
d d


19. A barometer tube reads 75 cm of Hg. If tube is gradually inclined at an angle of 30° with horizontal, keeping the
open end in the mercury container, then find the length of mercury column in the barometer tube
(1) 86.7 cm (2) 150 cm (3) 75 cm (4) 92.5 cm
Sol. Answer (2)
∵ We are not changing the atmospheric pressure, so
height of Hg from the surface should not change.
75
cos60
x
 
x 75 cm
60°
30°
 x = 150 cm

20. A metallic sphere weighing 3 kg in air is held by a string so as to be completely immersed in a liquid of relative
density 0.8. The relative density of metallic is 10. The tension in the string is
(1) 18.7 N (2) 42.5 N (3) 32.7 N (4) 27.6 N
Sol. Answer (4)
T
v,ML
FB
Mg
Where,
Force of Buoyancy
Density of liquid
Density of metal
Mass of sphere
Tension in string
B
L
M
F
M
T
⎧
⎪ ⎪
⎪ ⎪
⎨
 ⎪
⎪ 
⎪
⎪⎩
T = Mg – Buoyant Force
 T = MVg – LVg M
M
V
⎛ ⎞
 ⎜ ⎟
⎝ ⎠
= (M – L) Vg
3
(10 0.8) 10
10
   
 T = 9.2 × 3 = 27.6 N
21. A rectangular block is 10 cm  10 cm  15 cm in size is floating in water with 10 cm side vertical. If it floats with
15 cm side vertical, then the level of water will
(1) Rise (2) Fall
(3) Remain same (4) Change according to density of block
Sol. Answer (3)
Mass of block remains same, volume displaced of water will also remain same so level of water will not change.
22. Two cubical blocks identical in dimensions float in water in such a way that 1st block floats with half part immersed
in water and second block floats with 3/4 of its volume inside the water. The ratio of densities of blocks is
(1) 2 : 3 (2) 3 : 4 (3) 1 : 3 (4) 1 : 4
Sol. Answer (1)
2 1
3 1
4 2
V g V g      
1
2
Where,
Volume of block
density of liquid 1
density of liquid 2
V
⎧
⎪ ⎪
⎨
 ⎪
⎪ ⎩ 2
1
2
3



23. On putting a capillary tube in a pot filled with water, the level of water rises upto a height of
4 cm in the tube. If a tube of half the diameter is used instead, the water will rise to a height of nearly
(1) 2 cm (2) 4 cm (3) 8 cm (4) 11 cm
Sol. Answer (3)
For capillary tube
2T
h
r g



We can say
1 1
h or h
r d
 
So, 1 2
2 1
h d
h d


4
2
d
x d

 x = 8 cm
24. A flat plate of area 0.1 m2 is placed on a flat surface and is separated from it by a film of oil 10–5 m thick whose
coefficient of viscosity is 1.5 N sm–2. The force required to cause the plate to slide on the surface at constant
speed of 1 mm s–1 is
(1) 10 N (2) 15 N (3) 20 N (4) 25 N
Sol. Answer (2)
v
F A
l
 
Substituting values,
3
5
1 10
1.5 0.1
10



  
= 15 N
25. A ball of density  and radius r is dropped on the surface of a liquid of density  from certain height. If speed of ball
does not change even on entering in liquid and viscosity of liquid is , then the height from which ball dropped is
(1)
2
( )
2
9
r
g
⎡ ⎤  
⎢ ⎥
⎣ ⎦
(2)
2 2
2 ( )
9
g r  

(3)
2
2( )
9
gr  

(4)
2
2
( )
2
9
r
g
⎡ ⎤  
⎢ ⎥
⎢ ⎥⎣ ⎦
Sol. Answer (4)
The ball has already reached the magnitude of velocity which is equal to its terminal velocity in fluid.
2v gh {Velocity of body fallen form height 2h gh }
Terminal
2
2
( )
9
r
v g   

Equating both
2
2
2 ( )
9
r
gh g   


22
( )
2
9
r
h g
⎡ ⎤  
 ⎢ ⎥
⎢ ⎥⎣ ⎦

26. Viscous drag force depends on
(1) Size of body (2) Velocity with which it moves
(3) Viscosity of fluid (4) All of these
Sol. Answer (4)
V
F A
d
 
Where,
Drag Force
Viscosity of fluids
Area size of body
Velocity
F
A
V
⎧
⎪ ⎪⎪
 ⎨
⎪  
⎪
⎪⎩
27. The terminal velocity of a small sized spherical body of radius r falling vertically in a viscous liquid is given by the
proportionality
(1) 2
1
r
v  (2) v  r2 (3)
r
v
1
 (4) v  r
Sol. Answer (2)
2
2
[ ]
9T
r
v g   

So, vT  r2
28. A spherical ball is dropped in a long column of viscous liquid. The speed v of the ball varies as function of time as
(1)
v
t(0, 0)
(2)
v
t(0, 0)
(3)
v
t(0, 0)
(4)
v
t(0, 0)
Sol. Answer (2)
VTerm.
(0, 0) t
V
Velocity does not increases after terminal speed is achieved.
29. Which of the following is not the property of an ideal fluid?
(1) Fluid flow is irrotational (2) Fluid flow is streamline
(3) Fluid is incompressible (4) Fluid is viscous
Sol. Answer (4)
An ideal fluid is not viscous.

30. Water is flowing through a channel (lying in a vertical plane) as shown in the figure. Three sections A, B and C are
shown. Sections B and C have equal area of cross section. If PA, PB and PC are the pressures at A, B and C
respectively then
A
B
C
h
(1) PA > PB = PC (2) PA < PB < PC
(3) PA < PB = PC (4) PA > PB > PC
Sol. Answer (2)
Solution by using Bernoulli's principle and equation of continuity
Comparing points A and B
AAvA = ABvB {equation of continuity}
∵ AA < AB
vA > vB
2 21 1
2 2
A A B BP V gh P V gh         {Bernoulli's equation}
∵ vA > vB

2 21 1
2 2
A BV V  
 PA < PB ...(1)
Now comparing C and B
AB = AC  vB = vC [equation of continuity]
2 21 1
2 2
B B CC
P V gh P V gh        
 PB + ghB = PC + ghC
∵ hB > hC then ...(2)
PB < PC
Using (1) and (2)
We can say, PA < PB < PC
31. A liquid flows in the tube from left to right as shown in figure. A1 and A2 are the cross-sections of the portions of the
tube as shown. The ratio of speed
2
1
v
v
will be
A2 2,vA1v1
(1)
2
1
A
A
(2)
1
2
A
A
(3)
1
2
A
A
(4)
2
1
A
A

Sol. Answer (2)
By equation of continuity
v1A1 = v2A2
So,
1 2
2 1
v A
v A

32. Water ( = 1000 kg/m3) and kerosene ( = 800kg/m3) are filled in two identical cylindrical vessels. Both vessels
have small holes at their bottom. The speed of the water and kerosene coming out of their holes are v1 and v2
respectively. Select the correct alternative
(1) v1 = v2 (2) v1 = 0.8 v2 (3) 0.8 v1 = v2 (4) 21 8.0 vv 
Sol. Answer (1)
Velocity of efflux for small holes 2gh
Which clearly is independent of '' (density)
So, v1 = v2
33. Atank is filled with water to a height H.Ahole is made in one of the walls at a depth D below the water surface. The
distance x from the foot of the wall at which the stream of water coming out of the tank strikes the ground is given
by
(1) x = 2 [D (H – D)]1/2 (2) x = 2 (gD)1/2 (3) x = 2 [D (H + D)]1/2 (4) None of these
Sol. Answer (1)
Velocity of efflux 2gD v 
Say time taken by water to travel the vertical distance of (H – D) = 't'
Using 21
2
s ut at 
Where,
0
s H D
u
a g
⎧
⎪  ⎪
⎨
⎪
⎪ ⎩
D
v
x
H
We get,
2( )
' '
H D
t
g


Now, x = v × t
Substituting the values
2( )
2
H D
x gD
g

 
 x = 2[D (H – D)]1/2
34. A tank is filled with water and two holes A and B are made in it. For getting same range, ratio of
h/h is
A
B
(1) 2 (2)
2
1
(3)
3
1
(4) 1

Sol. Answer (4)
For hole 'A' For hole 'B'
Velocity of efflux 2 ( )g x h  Velocity of efflux 2gh
R = 2[(x + h')h]1/2 ...(1) R = 2[h'(x + h)]1/2 ...(2)
Equating (1) and (2)
We get
2[(x + h')h]1/2 = 2[h'(x + h)]1/2
 (x + h')h = h'(x + h)
 h = h'
 1
h
h


35. Water is filled in a tank upto 3 m height. The base of the tank is at height 1 m above the ground. What should be
the height of a hole made in it, so that water can be sprayed upto maximum horizontal distance on ground?
3 m
1 m
(1) 3 m from ground (2) 1.5 m from ground
(3) 1.5 m from base of tank (4) 2 m from ground
Sol. Answer (4)
Let height of hole from the base of container be h
Velocity of efflux 2 (3 )g h 
R = 2[(h + 1) (3 – h)]1/2 [R = Range proved in Q. 33]
 R = 2(– h2 + 2h + 3)1/2
2 2
dR
h
dh
  
If 0,
dR
dh
 then range would be more for corresponding height
So, 0 = – 2h + 2  h = 1
 Height from the ground = 1 + 1 = 2 m

36. Soap helps in cleaning clothes, because
(1) It attracts the dirt particles
(2) It decreases the surface tension of water
(3) It increases the cohesive force between water molecules
(4) It increases the angle of contact
Sol. Answer (2)
Soap helps cleaning clothes because, it decreases the surface tension of water thus water molecules penetrate
easily into dirt and oil.
37. On increasing temperature of a liquid, its surface tension generally
(1) Increases (2) Decreases
(3) Remains constant (4) First increases and then decreases
Sol. Answer (2)
On increasing the temperature energy increases hence surface tension decreases. Because surface tension is
nothing but some extra energy required by surface molecules to stay at the place.
38. The raincoats are made water proof by coating it with a material, which
(1) Absorb water (2) Increase surface tension of water
(3) Increase the angle of contact (4) Decreases the density of water
Sol. Answer (3)
Raincoats are coated with material which increase the angle of contact, so water does not penetrates inside the
layer.
39. An iron needle slowly placed on the surface of water floats because
(1) It displaces water more than its weight
(2) The density of material of needle is less than that of water
(3) Of surface tension
(4) Of its shape
Sol. Answer (3)
Mg
F F
F cosF cos
F sin
F sin
OO
Needle floats due to surface tension of water which balances the weight of needle.
In equilibrium 2F sin = mg

40. Two water droplets merge with each other to form a larger droplet. In this process
(1) Energy is liberated (2) Energy is absorbed
(3) Energy is neither liberated nor absorbed (4) Some mass is converted into energy
Sol. Answer (1)
Work is done when we break a drop into 'n' drops equal to 4R2(n1/3 – 1)
So energy will be liberated if we merge back those drops.
41. The radius of a soap bubble is r. The surface tension of soap solution is ‘S’. Keeping temperature constant, the radius
of the soap bubble is doubled. The energy necessary for this will be
(1) 24 r2S (2) 8 r2S (3) 16 r2S (4) 12 r2S
Sol. Answer (1)
Work done in making a soap bubble of radius r = 4 r2S × 2 = 8 r2S
 Energy of bubble = 8 r2S = Er
Multiply by 2 due
to two free surface
⎧ ⎫
⎨ ⎬
⎩ ⎭
Work done in making a 2r radius soap bubble = 4 (2r2) S × 2 = 32 r2S
 Energy of bubble = 32 r2S = E2r
So energy required to expand a bubble from r to 2r will be equal to E2r – Er
Substituting values
We get,
32 r2S – 8 r2S = 24 r2S
42. The surface tension of a liquid is 5 N/m. If a film is held on a ring of area 0.02 m2, its surface energy is about
(1) 5 × 10–2 J (2) 2.5 × 10–2 J (3) 2 × 10–1 J (4) 3 × 10–1 J
Sol. Answer (3)
Surface energy = surface tension × area of film × number of free surface
= 5 × 0.02 × 2
= 2 × 10–1 J
43. Two soap bubbles having radii 3 cm and 4 cm in vacuum, coalesce under isothermal conditions. The radius of the
new bubble is
(1) 1 cm (2) 5 cm (3) 7 cm (4) 3.5 cm
Sol. Answer (2)
Energy initial = Energy final
 8 (3)2S + 8  (4)2S = 8  (r)2S {Surface tension remains constant throughout process
 (3)2 + (4)2 = (r)2
 5 cm = r

44. The excess pressure in a soap bubble is double that in other one. The ratio of their volume is
(1) 1 : 2 (2) 1 : 8 (3) 1 : 4 (4) 1 : 1
Sol. Answer (2)
Excess pressure in soap bubble
4
=
S
R
Let for first bubble,
4S
P
R

Surface tension
Radius
S
R
⎧
⎨
⎩
For second bubble,
4
2
S
P
x

Surface tension
Radius
S
x
⎧
⎨
⎩
Substitute value of P
4 4
2
S S
R x
 

2
R
x 
Ratio of Radii
/ 2 1
2
R
R
 
So, Ratio of volume = (Ratio of Radii)3
3
1 1
2 8
⎛ ⎞
 ⎜ ⎟
⎝ ⎠
45. The work done to break a spherical drop of radius R in n drops of equal size is proportional to
(1) 1
1
2/3

n
(2) 1
1
1/3

n
(3) n1/3 – 1 (4) n4/3 – 1
Sol. Answer (3)
∵ Volume = constant
i.e., 3 34 4
3 3
R n r   
Radius of each new droplet 1/3
R
n

Work done to break into 'n' drops = S [n × 4r2 – 4R3]
= S × 4R2 [n1/3 – 1]
46. The kerosene oil rises up in the wick of a lamp
(1) Due to high surface tension of oil (2) Because the wick attract s the oil
(3) Because wick decreases the surface tension of oil (4) Due to capillaries formed in the wick
Sol. Answer (4)
Capillary action is responsible.
Wick has a lot of capillaries which help the oil rise.

47. Ploughing help to retain water by soil
(1) By creating capillaries (2) By breaking capillaries
(3) By turning the soil upside down (4) None of these
Sol. Answer (2)
By breaking capillaries as they do not allow water to seep inside.
48. A capillary tube of radius r is immersed in a liquid and mass of liquid, which rises up in it is M. If the radius of tube
is doubled, then the mass of liquid which will rise in capillary tube will be
(1) 2 M (2) M (3) M/2 (4) M/4
Sol. Answer (1)
M = v
 Mass  volume of tube
Mass  height × Area
2
Area
1
Height
Where radius of cube
r
r
r
⎧ 
⎪
⎪
⎨
⎪
⎪
⎩
2S
h
rgh
⎛ ⎞
⎜ ⎟
⎝ ⎠
 Mass 21
r
r
 
Mass  r
 If radius doubles, mass of liquid that rises up also doubles.
49. A massless inextensible string in the form of a loop is placed on a horizontal film of soap solution of surface
tension T. If film is pierced inside the loop and it convert into a circular loop of diameter d, then the tension
produced in string is
(1) Td (2) Td (3) d2T (4)
2
4
d T
Sol. Answer (1)
T cos  T cos 
T TT sin  T sin 
SF
 T cos  T cos 
T sin  T sin 

SF

By force balancing in vertical direction
SF = 2T sin 
SF = 2T 
is small
sin
⎧
⎨
 ⎩
∵
S × 2r × 2 = 2 × T × 
S × 2r = Tension radius
diameter
r
d
⎧
⎨
⎩S × d = Tension
Where,
Force due to surface tension
Tension in string
Small angle
Surface tension
FS
T
S
⎧
⎪ ⎪
⎪
⎨
⎪ 
⎪
⎪⎩
∵ S = T
So, Tension = Td

SECTION - B
Objective Type Questions
1. The atmospheric pressure at a place is 105 Pa. If tribromomethane (specific gravity = 2.9) be employed as
the barometric liquid, the barometric height is
(1) 3.52 m (2) 1.52 m (3) 4.52 m (4) 2.52 m
Sol. Answer (1)
1 atm 105 Pa = 76 cm of Hg
Density of Hg = 13.6
Hg × g × hHg = TBM × g × hTBM
Substituting values
density of Hg
height of Hg
density of
tribromomethane
height of tribromomethane
Hg
Hg
TBM
TBM
h
h
 ⎧
⎪
⎪
⎪
 ⎨
⎪
⎪
⎪ 
⎩13.6 × 76 = 2.9 × h
3.52 m = h
2. A large vessel of height H, is filled with a liquid of density , upto the brim . A small hole of radius r is made
at the side vertical face, close to the base. The horizontal force is required to stop the gushing of liquid is
(1) (gH)r2 (2) gH (3) gHr (4) gr2
Sol. Answer (1)
Pressure close to the base = gH
Force required = pressure × area of hole = gH(r2)
3. A vertical U-tube of uniform cross-section contains water in both the arms. A 10 cm glycerine column
(R.D. = 1.2) is added to one of the limbs. The level difference between the two free surfaces in the two limbs
will be
(1) 4 cm (2) 2 cm (3) 6 cm (4) 8 cm
Sol. Answer (2)
Let the difference between 2 limbs be x
Pressure on the line Y Y' should be same below both limbs, so
glycerine × g × h = water × g × (h + x)
 1.2 × 10 = 1 × (h + x)
x
P
 = 1.2
YY
P
h = 10 cm
 2 cm = x
4. The pressure at the bottom of a water tank is 4 P, where P is atmospheric pressure. If water is drawn out till
the water level decreases by
3
5
th, then pressure at the bottom of the tank is
(1)
3
8
P
(2)
7
6
P
(3)
11
5
P
(4)
9
4
P
Sol. Answer (3)
Let height of water in tank be h
So, 4P – P = wgh ...(1)

∵
3
5
water taken out
2
th
5
water is left to exert pressure
2
5
wP P gh   

2
3
5
P P P    [From eq. (1)]

11
5
P
P 
5. A air bubble rises from bottom of a lake to surface. If its radius increases by 200% and atmospheric pressure
is equal to water coloumn of height H, then depth of lake is
(1) 21 H (2) 8 H (3) 9 H (4) 26 H
Sol. Answer (4)
Let initial radius be = r
Final radius = r + 200% of r
= 3r
Atmospheric pressure = gH h
3r
r
(1)
P gH= 
P
P gH gh  = + (2)
Let depth of the lake be h
So, pressure at the bottom of lake = gH + gh
Using P1V1 = P2V2
3 34 4
(3 ) ( )
3 3
gH r gH gh r        
3 3 34 4 4
27 ( )
3 3 3
gH r gH r gh r         
Solving this equation we get
26 H = h
6. A piece of gold weighs 10 g in air and 9 g in water. What is the volume of cavity?
(Density of gold = 19.3 g cm–3)
(1) 0.182 cc (2) 0.282 cc (3) 0.382 cc (4) 0.482 cc
Sol. Answer (4)
Vg Vc
When dipped in water
Wapp = Wair – FB
 9 gm × g = 10 gm × g – FB
 1 × g = FB
Now (total volume displaced) × w × g = 1 × g
app
air
Where,
volume of cavity
volume of gold
9 gm
10 gm
force of buoyancy
density of water 1
density of gold 19.3
c
g
B
w
g
V
V
W
W
F
⎧
⎪ ⎪
⎪ 
⎪
⎪⎪
⎨
⎪
⎪ 
⎪
  ⎪
⎪
  ⎪⎩(Vc + Vg) × 1 = 1
Mass of gold in air 10
1 1 0.482 cc
19.3
c
g
V     


7. A block of ice floats in an oil in a vessel when the ice melts, the level of oil will
(1) Go up (2) Go down
(3) Remain same (4) Go up or down depending on quantity of ice
Sol. Answer (2)
Since block of ice is displacing some oils to stay afloat when the ice block melts level of oil will go down.
8. An object suspended by a wire stretches it by 10 mm. When object is immersed in a liquid the elongation in
wire reduces by
10
3
mm. The ratio of relative densities of the object and liquid is
(1) 3 : 1 (2) 1 : 3 (3) 1 : 2 (4) 2 : 1
Sol. Answer (1)
FL
L
AY
 
 Elongation  force and force is due to weight
Let density of liquid
Let density of object
Mass of object M
 ⎧
⎪
 ⎨
⎪ ⎩
So elongation  weight
L1  weight ...(1) {When not submerged in liquid}
L2  apparant weight ...(2) {When submerged in liquid}
Dividing (1) by (2)
10
10
10
3
Mg
Mg
Mg


 


1 1
1
1 1
3


 

Solving this we get
1
3



So relative densities of object () and liquid () is 3 : 1
9. Water flows in a stream line manner through a capillary tube of radius a. The pressure difference being P and the
rate of flow is Q. If the radius is reduced to
4
a
and the pressure is increased to 4P, then the rate of flow becomes
(1) 4Q (2)
2
Q
(3) Q (4)
64
Q
Sol. Answer (4)
Rate of flow  pressure difference × (radius)4
Q  P × a4
4
8
Pr
Q
L
⎧ ⎫⎪ ⎪
⎨ ⎬
⎪ ⎪⎩ ⎭
∵
So,
4
1 1 1
4
2 2 2
Q P a
Q P a

4
1
4
2
64
1
4
4
Q P a
Q a
P

 
⎛ ⎞
 ⎜ ⎟
⎝ ⎠
 1
2
64 64
Q Q
Q  

10. A vessel contain a liquid has a constant acceleration 19.6 m/s2 in horizontal direction. The free surface of water
get sloped with horizontal at angle
(1) 1 1
tan
2
 ⎡ ⎤
⎢ ⎥
⎣ ⎦
(2)
1 1
sin
3
 ⎡ ⎤
⎢ ⎥
⎣ ⎦
(3)
1
tan 2 ⎡ ⎤
⎣ ⎦ (4)
1 2
sin
5
 ⎡ ⎤
⎢ ⎥
⎣ ⎦
Sol. Answer (4)
19.6
tan 2
9.8
a
g
   
tan  = 2 a
g

a
 12 2
sin sin
5 5
 ⎡ ⎤
  ⇒   ⎢ ⎥
⎣ ⎦
11. A cylinder containing water, stands on a table of height H. A small hole is punched in the side of cylinder at
its base. The stream of water strikes the ground at a horizontal distance R from the table. Then the depth of
water in the cylinder is
(1) H (2) R (3) RH (4) R2/4H
Sol. Answer (4)
Let depth of water in cylinder be x
So velocity (v) of efflux 2gx
Time taken (t) by water to travel vertical distance of H
2H
g

x
H
R
v
 Range = v × t
2
2
H
R gx
g
 
Solving this we get
2
4
R
x
H

12. Air streams horizontally past an air plane. The speed over the top surface is 60 m/s and that under the bottom
surface is 45 m/s. The density of air is 1.293 kg/m3, then the difference in pressure is
(1) 1018 N/m2 (2) 516 N/m2 (3) 1140 N/m2 (4) 2250 N/m2
Sol. Answer (1)
Applying Bernoullis equation
P1 + gh +
1
2
v1
2 = P2 + gh +
1
2
v2
2
1
2
×  [v1
2 – v2
2] = P2 – P1 = P

1
2
× 1.293 [(60)2 – (45)2] = P
 1018 N/m2 P

13. Two water pipes P and Q having diameter 2 × 10–2 m and 4 × 10–2 m respectively are joined in series with
the main supply line of water. The velocity of water flowing in pipe P is
(1) Four times that of Q (2) Two times that of Q
(3)
1
2
times that of Q (4)
1
4
times that of Q
Sol. Answer (1)
Rate of flow through both pipes will be same
i.e., Q1 = Q2
1 2V V
t t

2 2
1 1 2 2r l r l
t t
 

1 2
Where andP Q
l l
V V
t t
⎛ ⎞
 ⎜ ⎟
⎝ ⎠

2 2
1 2
4 4
P Q
d d
V V
 
 

2
2
1
P Q
d
V V
d
⎛ ⎞
 ⎜ ⎟
⎝ ⎠

2
2
2
4 10
2 10
P QV V


⎛ ⎞
 ⎜ ⎟⎜ ⎟⎝ ⎠
 VP = 4VQ
14. At what speed, the velocity head of water is equal to pressure head of 40 cm of mercury?
(1) 2.8 m/s (2) 10.32 m/s (3) 5.6 m/s (4) 8.4 m/s
Sol. Answer (2)
1
2
water V2 = mercury gh
mercury
water
2V g h

   

40
2 13.6 9.8
100
   
 V = 10.32 m/s
15. If the terminal speed of a sphere of gold (density 19.5 kg/m3) is 0.2 m/s in a viscous liquid (density = 1.5
kg/m3), find the terminal speed of a sphere of silver (density = 10.5 kg/m3) of the same size in the same liquid.
(1) 0.2 m/s (2) 0.4 m/s (3) 0.1 m/s (4) 0.133 m/s

Sol. Answer (3)
2
terminal
2
( )
9
a
V g   

 VT  ( – )
Where
density of material
density of liquid
⎧
⎪
 ⎨
⎪ ⎩

1
2
gold liquid
silver liquid
T
T
V
V
  

  

0.2 19.5 1.5
10.5 1.5V



1
2
3
gold
3
liquid
3
silver
Given,
0.2 m/s
?
19.5 kg/m
1.5 kg/m
10.5 kg/m
T
T
V
V V
⎧
⎪ ⎪
⎪  
⎪
⎨
 ⎪
⎪
 ⎪
⎪
 ⎩
 V = 0.1 m/s
16. Three capillaries of length L,
2
L
and
3
L
are connected in series. Their radii are r,
2
r
and
3
r
respectively. Then
if stream-line flow is to be maintained and the pressure across the first capillary is P, then
(1) The pressure difference across the ends of second capillary is 8P
(2) The pressure difference across the third capillary is 43P
(3) The pressure difference across the ends of second capillary is 16P
(4) The pressure difference across the third capillary is 59P
Sol. Answer (1)
r r/2 r/3
L
P1
L/2 L/3
P2 P3
P1 = P (given)
∵ Rate of flow will be same across all pipes
So, pressure across the pipe 4
length
(radius)

4
rate of flow of liquid ( )
8
Q
Pr
Q
L
⎧
⎪
⎨ 
⎪ ⎩
 
 
4
1
2
4
/ 1
8/2
/2
L rP
P L
r
 
⎛ ⎞
⎜ ⎟
⎜ ⎟
⎝ ⎠
ThenP2 = 8P1
17. A large open tank has two holes in its wall. One is a square of side a at a depth x from the top and the other
is a circular hole of radius r at depth 4x from the top. When the tank is completely filled with water, the
quantities of water flowing out per second from both holes are the same. Then r is equal to
(1) 2a (2) a (3)
2
a

(4)
a


Sol. Answer (3)
Since quantities of water flowing out of both holes is same
 Area of hole × velocity of efflux = constant
So, A1 × V1 = A2 × V2
Substituting values.
1
1
2
2
Area of square hole
Velocity of efflux from square hole 2
Area of circular hole
Velocityof efflux from
circular hole 2 (4 )
A
V gx
A
V
g x
⎧
⎪
 ⎪
⎪
⎨
⎪ 
⎪
⎪ ⎩
2 2
2 8a gx r gx   
 a2 = 2r2

2
a
r

18. If T is the surface tension of a fluid, then the energy needed to break a liquid drop of radius R into 64 equal
drops is
(1) 6R2T (2) R2T (3) 12R2T (4) 8R2T
Sol. Answer (3)
Work done = surface tension × change in area
Since volume will remain equal
Let us assume radius of new drop = r each

3 34 4
64
3 3
R r   

4
R
r
W = T A
= T [n × 4r2 – 4R2]
2
2 2
64 4 4 12
4
R
T R R T
⎡ ⎤⎛ ⎞
      ⎢ ⎥⎜ ⎟
⎝ ⎠⎢ ⎥⎣ ⎦
19. The excess pressure inside a spherical drop of water is four times that of another drop. Then their respective
mass ratio is
(1) 1 : 16 (2) 1 : 64 (3) 1 : 4 (4) 1 : 8
Sol. Answer (2)
2T
P
R
 
Pressure
1
Radius

0
Where,
Excess pressure
Surface tension
Radius
P
T
R
⎧
⎪ ⎪
⎨
⎪
⎪ ⎩
 1 2
2 1
P R
P R




 2
1
4 RP
P R

 4R1 = R2
 1
2
1
4
R
R

M = V × 
And V  R3  M  R3
Where,
Mass
Volume
Density
M
V
⎧
⎪ ⎪
⎨
⎪
⎪ ⎩
 is same for both
M  R3
So,
3 3
1 1
2 2
1 1
4 64
M R
M R
⎛ ⎞ ⎛ ⎞
  ⎜ ⎟ ⎜ ⎟
⎝ ⎠⎝ ⎠
20. The work done in blowing a soap bubble of 10 cm radius is (surface tension of soap solution is
0.03 N/m).
(1) 37.68 × 10–4 J (2) 75.36 × 10–4 J (3) 126.82 × 10–4 J (4) 75.36 × 10–3 J
Sol. Answer (2)
Work done = surface tension × change in area × number of free surfaces = S × A × 2
= 0.03 × 4 × (10 × 10–2)2 × 2
= 75.36 × 10–4 J
21. A glass capillary tube of inner diameter 0.28 mm is lowered vertically into water in a vessel. The pressure to
be applied on the water in the capillary tube so that water level in the tube is same as that in the vessel is
(surface tension of water = 0.07 N/m and atmospheric pressure = 105 N/m2).
(1) 103 (2) 99 × 103 (3) 100 × 103 (4) 101 × 103
Sol. Answer (4)
Height of liquid in capillary
2T
h
r g
 

Pressure we need to apply = gh + P0
0
Where,
Surface tension
Radius of capillary
Density of liquid
Atmospheric pressure
T
r
P
⎧
⎪ ⎪⎪
⎨
⎪ 
⎪
⎪⎩
Substitute value of h
0 0 0
2 2 4T T T
P g P P P
r g r d
       

 03
4 0.07
(0.28 10 )
P P

 

= 1000 Nm–2 + 105 Nm–2
Given,
0.07 N/m
0.28 mm
T
d
⎧
⎪
⎨
⎪ ⎩
 P = (103 + 105) Nm–2 = 101 × 103 Nm–2
22. Water rises to a height of 10 cm in a capillary tube and mercury falls to a depth of 3.42 cm in the same
capillary tube. If the density of mercury is 13.6 kg/m3 and angle of contact is 135°, the ratio of surface tension
for water and mercury is (angle of contact for water and glass is 0°).
(1) 1 : 0.5 (2) 1 : 3 (3) 1 : 6.5 (4) 1.5 : 1

Sol. Answer (3)
2 cosT
h
r g



For water,
2 cos0
10 cm
1
wT
r g
  

 
...(1) {Tw – Surface tension of water
For mercury,
2 cos135
3.42 cm
13.6
MT
r g
  
 
 
...(2) {TM – Surface tension of mercury
Dividing Eqn (1) by (2)
2 1 13.610
13.42
1 2
2
w
M
T r g
r g T
    


    

10
2 13.6
3.42
w
M
T
T
  

10
3.42 1.41 13.6
w
M
T
T

 

1
6.5
w
M
T
T

23. A spherical drop of water has 1 mm radius. If the surface tension of water is 75 × 10–3 N/m, then difference
of pressure between inside and outside of the drop is
(1) 35 N/m2 (2) 70 N/m2 (3) 140 N/m2 (4) 150 N/m2
Sol. Answer (4)
Excess pressure
2T
R

surface tension
radius
T
R
⎧
⎨
⎩
3
3
2 75 10
1 10


 


= 150 N/m2
24. A capillary tube is dipped in water and it is 20 cm outside water. The water rises upto 8 cm. If the entire
arrangement is put in freely falling elevator the length of water column in the capillary tube will be
(1) 20 cm (2) 4 cm (3) 10 cm (4) 8 cm
Sol. Answer (1)
If entire arrangement is in free fall then the weight of water in capillary will be balanced by pseudo force which
would be equal to the weight of water.
Hence, surface tension has no weight to balance so full capillary will be filled with water.

25. If the excess pressure inside a soap bubble is balanced by an oil column of height 2 mm, then the surface
tension of soap solution will be (r = 1 cm, density of oil = 0.8 g/cm3)
(1) 3.9 N/m (2) 3.9 × 10–2 N/m (3) 3.9 × 10–3 N/m (4) 3.9 × 10–1 N/m
Sol. Answer (2)
Pressure due to oil column = oil × g × hoil
3
3
2 3
0.08 10
9.8 2 10 15.68
(10 )




    
Now, excess pressure = pressure due to oil column

4
15.68
T
R

 2
4
15.68
1 10
T




 T = 3.92 × 10–2 N/m
26. There is small hole in a hollow sphere. The water enters in it when it is taken to a depth of 40 cm under water.
The surface tension of water is 0.07 N/m. The diameter of hole is
(1) 7 mm (2) 0.07 mm (3) 0.0007 mm (4) 0.7 m
Sol. Answer (2)
Let take g = 10 m/s2
For water to enter the sphere, pressure required is = gh
40
1 10 1000
100
    ( = 1000 kg/m3)
R
= 4000 2
N
m
= excess pressure
Let the hole have radius = R
Excess pressure
2T
R
 [One surface air, one surface water]

2 0.07
4000
R


 2R = 0.07 × 10–3 m
 d = 0.07 mm
27. Two equal drops are falling through air with a steady velocity of 5 cm/second. If two drops coalesce, then new
terminal velocity will be
(1) 5 × (4)1/3 cm/s (2) 5 2 cm/s (3)
5
cm/s
2
(4) 5 × 2 cm/s
Sol. Answer (1)
VTerminal  r2
If initial radius = r, let new radius = R

Then 2 ×
4
3
r3 =
4
3
R3
 (2)1/3 r = R
 VT  R2
 (2)2/3 r2 (For bigger drops)
2
smaller drop
2/3 2
bigger drop (2)
T
T
V r
V r

 2/3
5 1
(2)x

 5 × (2)2/3 = x
 5 × (4)1/3 cm/s = x
28. A small drop of water falls from rest through a large height h in air; the final velocity is
(1) Proportional to h (2) Proportional to h
(3) Inversely proportional to h (4) Almost independent of h
Sol. Answer (4)
Since drop is falling from a large height it achieves its terminal velocity and then there is no further increase
in velocity so v is independent of 'h' if 'h' is very large.
29. A spring balance reads 200 gF when carrying a lump of lead in air. If the lead is now immersed with half of
its volume in brine solution, what will be the new reading of the spring balance? specific gravity of lead and
brine are 11.4 and 1.1 respectively
(1) 190.4 gF (2) 180.4 gF (3) 210 gF (4) 170.4 gF
Sol. Answer (1)
W' = W – FB
2
v
v g g   
1
2
v g
⎛ ⎞
  ⎜ ⎟⎝ ⎠
Where,
apparent weight
read weight = actual weight of body in vaccum
density of solution (1.1)
densityof material (11.4)
W
W
⎧
⎪  ⎪⎪
⎨
⎪ 
⎪
 ⎪⎩
1.1
200 1
11.4 2
W
⎛ ⎞
  ⎜ ⎟⎝ ⎠
= 190.35 gF
30. A liquid mixture of volume V, has two liquids as its ingredients with densities  and . If density of the mixture
is , then mass of the first liquid in mixture is
(1)
[ 1]
[ ]
V  
   
(2)
[ ]
[ ]
V   
  
(3)
( )V   
  
(4)
[1 ]
[ ]
V  
   

Sol. Answer (3)
Let mass of liquid with density  = M1
Let mass of liquid with density  = M2
Total volume = V
Net density of mixture = 
Now,
Total mass = M1 + M2
 V = M1 + M2
Total mass
V
⎡ ⎤
 ⎢ ⎥
⎣ ⎦
∵
 M2 = V – M1 ...(1)
Now,
1 2
1 2
( )Total mass
Total volume
M M
M M

  
⎛ ⎞⎛ ⎞
 ⎜ ⎟⎜ ⎟ ⎝ ⎠ ⎝ ⎠
Substituting value of M2 from equation (1)
1 1
1 1
( )
( )
M V M
M V M
  
 
 

 
Solving this we get
1
( )V
M
   

  
SECTION - C
Previous Years Questions
1. The neck and bottom of a bottle are 3 cm and 15 cm in radius respectively. If the cork is pressed with a force
12 N in the neck of the bottle, then force exerted on the bottom of the bottle is
(1) 30 N (2) 150 N (3) 300 N (4) 600 N
Sol. Answer (3)
Pressure applied on 1 point in a liquid spreads equally
So let P1 be pressure at neck, P2 be pressure at bottom
P1 = P2
 1 2
1 2
F F F
P
A A A
⎡ ⎤
 ⎢ ⎥
⎣ ⎦
∵
15 cm
P2
3 cm
P1

212
9 225
F

 
 300 N = F2

2. A liquid X of density 3.36 g cm–3 is poured in a U-tube, which contains Hg. Another liquid Y is poured in left
arm with height 8 cm, upper levels of X and Y are same what is density of Y?
Y
10 cm
8 cm
X
(1) 0.8 gcc–1 (2) 1.2 gcc–1 (3) 1.4 gcc–1 (4) 1.6 gcc–1
Sol. Answer (1)
Pressure at 1 and 2 will be same
X gHX = Y gHY + Hg g × 2
 3.36 × 10 = Y × 8 + 13.6 × 2
Y
10 cm
8 cm
X
12
2 cm
Solving this we get
Y = 0.8 g cc–1
3. A wooden ball of density D is immersed in water of density d to a depth h below the surface of water and
then released. Upto what height will the ball jump out of water?
(1)
d
h
D
(2) 1
⎛ ⎞
⎜ ⎟
⎝ ⎠
d
h
D
(3) h (4) Zero
Sol. Answer (2)
Force acting on ball at depth 'h' (i.e. apparent weight)
F = Vg [d – D]
Acceleration
[ ]
( )
Vg d D
a
VD

 [Mass = volume × density]
Velocity 2ah v  [Using v2 – u2 = 2as]
h' (height above water)
2
2 [ ]
1
2 2
v Vg d D h d
h
g gVD D
  ⎡ ⎤
   ⎢ ⎥ ⎣ ⎦
h
h
VDg
v
Vdg
4. A piece of solid weighs 120 g in air, 80 g in water and 60 g in a liquid. The relative density of the solid and
that of the liquid are respectively
(1) 3, 2 (2)
3
2,
4
(3)
3
, 2
2
(4)
3
3,
2

Sol. Answer (4)
1w w
⎡ ⎤
  ⎢ ⎥⎣ ⎦
Where,
density of liquid
density of body
⎧
⎪
 ⎨
⎪ ⎩Inside water
 water
solid
80 120 1
⎡ ⎤
 ⎢ ⎥
⎣ ⎦
 solid = 3 [∵ water = 1]
Inside liquid
liquid
solid
60 120 1
⎡ ⎤
 ⎢ ⎥
⎣ ⎦
Using solid = 3
We get liquid
3
2
 
5. A solid sphere of volume V and density ρ floats at the interface of two immiscible liquids of densities ρ1 and
ρ2 respectively. If ρ1 < ρ < ρ2, then the ratio of volume of the parts of the sphere in upper and lower liquids is
(1)
2
1
  
   (2)
1
2
  
   (3)
2
1
  
   (4)
1 2 

Sol. Answer (1)
ρ1 < ρ < ρ2 (given)
Let volume of sphere in lower liquid = x
Force of buoyancy by lower liquid = ρ2xg
Force of buoyancy by upper liquid = ρ1(V – x)g
Force of gravity on sphere = Mg = Vρg
Mg
2 xg

1( – )V x g
1
2
Balancing all the forces for vertical equilibrium
We get
Vρg = ρ1(V – x) g + ρ2xg
Solving this we get
1
2 1
( )
( )
V
x
  

  
So 2
1
V x
x
  

  
Where,
volume of sphere in upper liquid
volume of sphere in lower liquid
V x
x
⎧
⎪
 ⎨
⎪ ⎩
6. Ice pieces are floating in a beaker A containing water and also in a beaker B containing miscible liquid of
specific gravity 1.2. When ice melts, the level of
(1) Water increases in A (2) Water decreases in A
(3) Liquid in B decreases (4) Liquid in B increases

Sol. Answer (4)
For beaker 'A'
Ice is floating in water
ice vice g = water vwater displaced g
∵ ice water
So we can say
vice vwater displaced
So after the ice melts the level of water will not change.
For beaker 'B'
Ice is floating in liquid with density 1.2
clearly liquid > ice
So from above analogy
vice > vliquid displaced
So when ice melts the level in beaker 'B' increases.
7. A vessel contains oil (density 0.8 g cm–3) over mercury (density 13.6 g cm–3). A homogenous sphere floats
with half volume immersed in mercury and the other half in oil. The density of the material of the sphere in g
cm–3 is
(1) 12.8 (2) 7.2 (3) 6.4 (4) 3.3
Sol. Answer (2)
Let density of sphere be 
And volume be v
Balancing forces for vertical equilibrium
oil
2 2
HgVg Vg
Vg
 
  
oil
Mercury
Vg 2HgB Hg
V
F g 
oil oil
2B
V
F g 

13.6 0.8
2 2
  
  = 7.2 g cm–3
8. Two solid pieces, one of steel and the other of aluminium when immersed completely in water have equal
weights. When the solid pieces are weighed in air
(1) The weight of aluminium is half the weight of steel (2) Steel peice will weigh more
(3) They have the same weight (4) Aluminium piece will weigh more
Sol. Answer (4)
Apparent weight = weight in air – FBuoyancy
∵ Apparent weight of steel and aluminium is same
So weight of aluminium – FB on Aluminium = weight of steel – FB on steel ...(1)
Al VAl g – water VAl g = steel Vsteel g – water Vsteel g
steel > Al and water = 1

So (Al – 1) VAl = (steel – 1) Vsteel
∵ steel > Al
(steel – 1) > (Al – 1)
So VAl > Vsteel
Also water VAl g > water Vsteel g
 Force of buoyancy on Aluminium > Force of buoyancy on steel.
Using this condition in equation (1)
We get,
weight of Aluminium – weight of steel > 0
 weight of Aluminium > weight of steel
9. A piece of wood is floating in water. When the temperature of water rises, the apparent weight of the wood
will
(1) Increase (2) Decrease
(3) May increase or decrease (4) Remain same
Sol. Answer (3)
When temperature of water is raised from 0 to 4°C, its density increases and after 4°C density decreases and
apparent weight  FBuoyancy  density of water
So apparent weight may increase or decrease.
10. A wooden block, with a coin placed on its top, floats in water as shown in the figure. The distances h and l
are shown there. After some time, the coin falls into the water, then
l
h
Coin
(1) Both l and h increase (2) Both l and h decrease
(3) l decreases and h increases (4) l increases and h decreases
Sol. Answer (2)
When coin falls into water block has to displace lesser volume to stay afloat.
Implies that block will go up and water will go down.
Hence both l and h will decrease.
11. An iceberg is floating in water. The density of ice in the iceberg is 917 kg m–3 and the density of water is
1024 kg m–3. What percentage fraction of the iceberg would be visible?
(1) 5% (2) 10% (3) 12% (4) 8%
Sol. Answer (2)
ice × volume of ice × g = water × volume of ice inside water × g
917 × volume of ice = 1024 × volume of ice inside water

Let volume of ice = V
 % Volume visible
volume inside water
100
V
V

 
917
1024 100
V
V
V
⎛ ⎞
⎜ ⎟
 ⎜ ⎟
⎝ ⎠
1024 917
1024
100
V
V
⎛ ⎞
⎜ ⎟
⎝ ⎠ 
= 10%
12. A piece of wax wieghs 18.03 g in air. A piece of metal is found to weigh 17.3 g in water. It is tied to the wax
and both together weigh 15.23 g in water. Then, the specific gravity of wax is
(1)
18.03
17.03
(2)
17.03
18.03
(3)
18.03
19.83
(4)
15.03
17.03
Sol. Answer (3)
Weight of wax in air = 18.03 g
Apparent weight of metal in water = 17.3 g
Apparent weight = weight in air – water Vmetal g
So weight of metal in air = apparent weight + Vmetal g [∵ water = 1]
= 17.3 + Vmetal × g
When wax and metal are tied together
Total weight in air = 18.03 + 17.3 + Vmetal × g
And apparent weight in water = 15.23 = weight in air – water Vwax g – water Vmetal g
 15.23 = 18.03 + 17.3 + Vmetal g – Vwax g – Vmetal g
 Vwax g = 20.1

Mass of wax
20.1
density
M
g
V
⎡ ⎤
   ⎢ ⎥
⎣ ⎦
∵

18.03 weight
20.1 Massg
g g
⎡ ⎤
  ⎢ ⎥  ⎣ ⎦
So specific gravity of wax
18.03
0.897 0.9
20.1
  ∼

18.03
0.9
19.83
⎛ ⎞
⎜ ⎟
⎝ ⎠

13. Eight equal drops of water are falling through air with a steady velocity of 10 cm–1. If the drops combine to
form a single drop big in size, then the terminal velocity of this big drop is
(1) 80 cms–1 (2) 30 cms–1 (3) 10 cms–1 (4) 40 cms–1
Sol. Answer (4)
Let radius of smaller drops be r, and bigger be R
When 8 such drops combine to form a bigger drop the total volume of water remains same
So, 3 34 4
8
3 3
r R   
 2r = R
And we know,
VTerminal  r2

2 2
smaller
2 2
bigger 4
T
T
V r r
V R r
 

bigger
10 1
4TV
 [∵ VT smaller = 10 cms–1 (given)]
 VT bigger = 40 cms–1
14. A small spherical ball falling through a viscous medium of negligible density has terminal velocity . Another
ball of the same mass but of radius twice that of the earlier falling through the same viscous medium will have
terminal velocity
(1) ν (2)
4

(3)
2

(4) 4ν
Sol. Answer (4)
vTerminal  r2
1
2
2
1
2
2
T
T
v r
v r

Substituting values
1
2
1
2
Where,
2
?
T
T
v
r r
r r
v
⎧
⎪
 ⎪
⎪
⎨
⎪ 
⎪
⎪ ⎩
2
2
2
4T
r
v r


 vT2
= 4
15. Streamline flow is more likely for liquid with
(1) High density and low viscosity (2) Low density and high viscosity
(3) High density and high viscosity (4) Low density and low viscosity
Sol. Answer (2)
Streamline flow is more likely for liquid with low density and high viscosity.

16. An air bubble of radius 10–2 m is rising up at a steady rate of 2 × 10–3 ms–1 through a liquid of density 1.5 × 103
kg m–3, the coefficient of viscosity neglecting the density of air, will be (g = 10 ms–2)
(1) 23.2 units (2) 83.5 units (3) 334 units (4) 167 units
Sol. Answer (4)
2
2
( )
9
T
a
V g   

Substituting values
4 3
3 2 10 10 1.5 10
2 10
9

    
 

3 1
2
3 3
2
Where given is
2 10 ms
10 m
1.5 10 kg m
0
10 ms
TV
a
g
 



⎧
⎪
 ⎪
⎪
⎪
⎨
  ⎪
⎪⎪
⎪ ⎩
∼
   167 units
17. The flow of liquid is laminar or streamline is determined by
(1) Rate of flow of liquid (2) Density of fluid
(3) Radius of the tube (4) Coefficient of viscosity of liquid
Sol. Answer (1)
It is decided by rate of flow of liquid
Given by Reynolds number
vd
n


18. A boat carrying a number of large stones is floating in a water tank. What would happen to the water level, if
a few stones are unloaded into water?
(1) Rises
(2) Falls
(3) Remains unchanged
(4) Rises till half the number of stones are unloaded and then begins to fall
Sol. Answer (2)
Previously when stones are on the boat they are increasing the weight on the boat and to balance this weight
boat needs to generate buoyancy force by displacing more water, but when stones are removed the boat starts
displacing less amount of water hence the level of water in tank falls.
Decreases in level
19. The velocity of a small ball of mass M and density d1 when dropped in a container filled with glycerine becomes
constant after some time. If the density of glycerine is d2, the viscous force acting on the ball is
(1) 2
1
1
⎛ ⎞
⎜ ⎟
⎝ ⎠
d
Mg
d
(2) 1
2
d
Mg
d
(3) mg(d1 – d2) (4) mgd1d2
Sol. Answer (1)
2 2
1 2 1
1 1
1 1v v
d d
F mg F vd g vd g vd g mg
d d
⎛ ⎞ ⎛ ⎞
       ⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
mg
Fv
FB

20. There are two holes one each along the opposite sides of a wide rectangular tank. The cross-section of each
hole is 0.01 m2 and the vertical distance between the holes is one metre. The tank is filled with water. The
net force on the tank in newton when the water flows out of the holes is (density of water = 1000 kgm–3)
(1) 100 (2) 200 (3) 300 (4) 400
Sol. Answer (2)
Net force = F2 – F1
= Av2
2 – Av2
1
2v v
F ma v Ah A v
t t
⎛ ⎞ ⎛ ⎞
      ⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
= 2g (h + 1) A – 2ghA 2v gx⎡ ⎤⎣ ⎦
h
1 m
v1
v2
= 2gA
= 1000 × 10 × 0.01 × 2
= 200 N
21. A hole is made at the bottom of the tank filled with water (density 1000 kg/m3). If the total pressure at the
bottom of the tank is 3 atm (1 atm = 105 N/m2), then the velocity of efflux is
(1) 200 m/s (2) 400 m/s (3) 500 m/s (4) 800 m/s
Sol. Answer (2)
Apply Bernoulli's theorem
2
Total pressure
1
constant
2
P gH v    
Given
Total pressure 3 atm
⎧
⎨
⎩
At point 1, 3 atm + 0 = constant ...(i)
At point 2, 1 atm +
1
2
v2 = constant ...(ii)
1 2
v
Equate (i) and (ii)
3 = 1 +
1
2
v2 [Use  = 1000 and 1 atm = 105 N/m2]
We get,
400 m/sv 
22. A horizontal pipe line carries water in stremline flow. At a point where the cross-sectional area is 10 cm2 the
water velocity is 1 ms–1 and pressure is 2000 Pa. The pressure of water at another point where the cross-
sectional area is 5 cm2, is
(1) 200 Pa (2) 400 Pa (3) 500 Pa (4) 800 Pa
Sol. Answer (3)
10 cm2
5 cm2
v1 ms
–1
P = 2000 Pa

A1V1 = A2V2 (equation of continuity)
10 × 1 = 5 × v
So, v = 2 ms–1
Apply Bernoulli theorem at both the points
2000 +
1
2
× 1000 × 12 = P +
1
2
× 1000 × 4
 P = 500 Pa
23. Arectangular vessel when full of water, takes 10 min to be emptied through an orifice in its bottom. How much time
will it take to be emptied when half filled with water?
(1) 9 min (2) 7 min (3) 5 min (4) 3 min
Sol. Answer (2)
Let time taken by height 'x' to get reduced by dx = dt

volume
efflux speed 2
A dx
dt
gx

  {A is area of cross-section}
0 0 2
T h A dx
dt
g x
∫ ∫

2A h
T
a g

 T h
A
x
dx
a
So we can use
11
2 2
hT
T h

10 min
min / 2
h
t h


10
7 min
2
t  ∼
24. A metal plate of area 103 cm2 rests on a layer of oil 6 mm thick. A tangential force 10–2 N is applied on it to
move it with a constant velocity of 6 cms–1. The coefficient of viscosity of the liquid is
(1) 0.1 poise (2) 0.5 poise (3) 0.7 poise (4) 0.9 poise
Sol. Answer (1)
F = A
v
d

2
2 3 4
3
6 10
10 (10 10 ) 0.01poiseuille
6 10

 


    

Where,
Force
Coefficient of viscosity
Area
Velocity
Thickness of layer
F
A
v
d
⎧
⎪ ⎪
⎪ ⎪
⎨
⎪
⎪ 
⎪
⎪⎩= 0.1 poise

25. With an increase in temperature, surface tension of liquid (except molten copper and cadmium)
(1) Increases (2) Remain same
(3) Decreases (4) First decreases then increases
Sol. Answer (3)
When we increase the temperature, we are providing energy to the molecules. This increase in potential energy
causes the surface energy to drop and become less negative hence decreasing surface tension because surface
tension is nothing but surface energy per unit area.
26. Determine the energy stored in the surface of a soap bubble of radius 2.1 cm if its tension is 4.5  10–2 Nm–1.
(1) 8 mJ (2) 2.46 mJ (3) 4.93 × 10–4 J (4) None of these
Sol. Answer (3)
Energy = surface tension × surface area × number of free surfaces
= (4.5 × 10–2) × 4 × (2.1 × 10–2) × 2
= 4.98 × 10–4 J
27. A mercury drop of radius 1.0 cm is sprayed into 106 droplets of equal sizes. The energy expended in this
process is (surface tension of mercury is equal to 32 × 10–2 Nm–1)
(1) 3.98 × 10–4 J (2) 8.46 × 10–4 J (3) 3.98 × 10–2 J (4) 8.46 × 10–2 J
Sol. Answer (3)
Energy expended = surface tension × increase in area (Formulae)
So, volume initially = volume of 106 drops

3
6 34 1 4
10
3 100 3
r
⎛ ⎞
   ⎜ ⎟
⎝ ⎠

1/3
3
6
1 1
100 10
r
⎡ ⎤⎛ ⎞
 ⎢ ⎥⎜ ⎟
⎝ ⎠⎢ ⎥⎣ ⎦
[Let radius of small drops = r]
 10–4 m = r
So increase in surface area
2 2
61 1
4 10
10000 100
⎡ ⎤⎛ ⎞ ⎛ ⎞
   ⎢ ⎥⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠⎢ ⎥⎣ ⎦
1 1
4
100 10000
⎡ ⎤
  ⎢ ⎥
⎣ ⎦
 4 0.99
100
A

 
Using this value in formulae
Energy 2 4 0.99
32 10
100
 
   [∵ Surface tension = 32 × 10–2 (given)]
= 3.98 × 10–2 J

28. When a glass capillary tube of radius 0.015 cm is dipped in water, the water rises to a height of
15 cm within it. Assuming contact angle between water and glass to be 0°, the surface tension of water is
[ρwater = 1000 kg m–3, g = 9.81 ms–2]
(1) 0.11 Nm–1 (2) 0.7 Nm–1 (3) 0.072 Nm–1 (4) None of these
Sol. Answer (1)
2 cosS
h
rg



Substituting values
15 2 1 100
100 1000 0.015 9.81
S  

  2
Where,
surface tension ?
height of water in capillary 15 cm
radius of capillary 0.015 cm
angle of contact 0
9.8 ms
S
h
r
g 
⎧
⎪  ⎪
⎪  ⎪
⎨
 ⎪
⎪   
⎪
⎪⎩
 S = 0.11 Nm–1
29. A liquid does not wet the sides of a solid, if the angle of contact is
(1) Obtuse (2) 90° (3) Acute (4) Zero
Sol. Answer (1)
Solid will not get wet if the liquid has high surface tension (example mercury) and liquids with high surface
tension have obtuse angle of contact.
30. Two drops of equal radius coalesce to form a bigger drop. What is ratio of surface energy of bigger drop to a
smaller one?
(1) 21/2 : 1 (2) 1 : 1 (3) 22/3 : 1 (4) None of these
Sol. Answer (3)
Surface energy = surface tension × surface area
Let the radius of smaller drops be r
And that of bigger drop be R
Then ratio of surface energies = ratio of surface area [∵ Surface tension is same for both]
= 4R2 : 4r2
= R2 : r2 ...(1)
∵ 2 smaller drops are forming 1 big drop so
3 34 4
2
3 3
r R   
So, 21/3r = R ...(2)
 Using 1 and 2 we can say that ratio of surface energies = 22/3r2 : r2 = 22/3 : 1
31. The excess pressure inside a spherical drop of water is frour times that of another drop. Then their respective
mass ratio is
(1) 1 : 16 (2) 8 : 1 (3) 1 : 4 (4) 1 : 64

Sol. Answer (4)
Excess pressure
2T
r
 {Where, r = radius of drop
1
P
r
 
1 2 1
2 1 2
4 1
1 4
P r P
P r P
⎡ ⎤ 
  ⎢ ⎥
 ⎣ ⎦
∵
V  r3
3 3
1 1
2 2
1 1
4 64
V r
V r
⎛ ⎞ ⎛ ⎞
  ⎜ ⎟ ⎜ ⎟
⎝ ⎠⎝ ⎠
 M = V ×  then
1 1
2 2
1
64
M V
M V
 
32. A balloon with mass m is descending down with an acceleration a (where a < g). How much mass should be
removed from it so that it starts moving up with an acceleration a?
(1)
2ma
g a
(2)
2ma
g a
(3)
ma
g a
(4)
ma
g a
Sol. Answer (1)
a
N
m
mg
mg – N = ma ..... (1)
Let m mass remove from ballon
a
N
( )m –m
( )m –m g
N – (m – m)g = (m – m)a ..... (2)
After addition of equation (1) & (2), then
2
'
ma
m
g a



33. A certain number of spherical drops of a liquid of radius r coalesce to form a single drop of radius R and volume V.
If T is the surface tension of the liquid, then
(1) Energy =
1 1
4VT
r R
⎛ ⎞
⎜ ⎟
⎝ ⎠
is released (2) Energy =
1 1
3VT
r R
⎛ ⎞
⎜ ⎟
⎝ ⎠
is absorbed
(3) Energy =
1 1
3VT
r R
⎛ ⎞
⎜ ⎟
⎝ ⎠
is released (4) Energy is neither released nor absorbed
Sol. Answer (3)
Let n drops of radius r coaslece to form a big drop of radius R
3 34 4
3 3
n r R V    
n × 4r3 = 4R3 = 3V ...(1)
Energy = T.A
= T [n × 4r2 – 4R2]
=
3 3
4 4n r R
T
r R
⎡ ⎤  
⎢ ⎥
⎢ ⎥⎣ ⎦
=
3 3V V
T
r R
⎡ ⎤
⎢ ⎥
⎣ ⎦
=
1 1
3VT
r R
⎡ ⎤
⎢ ⎥
⎣ ⎦
34. A wind with speed 40 m/s blows parallel to the roof of a house. The area of the roof is 250 m2. Assuming
that the pressure inside the house is atmospheric pressure, the force exerted by the wind on the roof and the
direction of the force will be (Pair = 1.2 kg/m3)
(1) 2.4 × 105 N, downwards (2) 4.8 × 105 N, downwards
(3) 4.8 × 105 N, upwards (4) 2.4 × 105 N, upwards
Sol. Answer (4)
2 2
1 1 2 2
1 1
2 2
P v P v    
 2 2
2 1 1 2
1
2
v v P P   
 2 2
2 1
1
2
F
v v
A
  
V2 = 40 m/s
P =1 P0
V = 01
P = P2
 2 21
1.2 40 0
2 250
F
   
F = 2.4 × 105 N, upward (because P1 > P2)

SECTION - D
Assertion - Reason Type Questions
1. A : Hydraulic lift is based on Pascal’s Law.
R : Hydrostatic pressure is a scalar quantity
Sol. Answer (2)
A : is true
R : is true
But reason is not the correct explanation. Correct explanation is, change in pressure is transferred, undiminished
from one point to the other.
2. A : The apparent weight of a body floating on the surface of a liquid is zero.
R : The net force on a body floating on the surface of a liquid is zero.
Sol. Answer (1)
A : is true
R : is true
And reason is also the correct explanation.
3. A : It is better to wash cloths in hot water than cold water.
R : On increasing temperature surface tension of water decreases.
Sol. Answer (1)
A : is true
R : is true and correct explanation.
4. A : The impurities added to water may increase or decrease surface tension.
R : The change in surface tension depends on the nature of impurities.
Sol. Answer (1)
A is true, R is true and correct explanation.
5. A : On increasing temperature the angle of contact generally decreases.
R : With rise in temperature, the surface tension of liquid increases.
Sol. Answer (4)
A : is false : Angle of contact increases with increase in temperature.
R : is false : Surface tension decreases with rise in temperature.
6. A : If air blows over the roof of a house, the force on the roof is upwards.
R : When air blows over the roof, the pressure over it from out side decreases.
Sol. Answer (1)
A is true, R is true and correct explanation.

7. A : When rain drops fall through air some distance, they attain a constant velocity.
R : The viscous drag of air just balances the weight of rain drops.
Sol. Answer (1)
A : is true
8. A : Bernoulli’s theorem holds good only for non-viscous and incompressible liquid.
R : Bernoulli’s theorem is based on the conservation of energy.
Sol. Answer (2)
A : is true
R : is true
Correct explanation of 'A' is Bernoulli's equation does not take into account the elastic energy of the fluids.
9. A : At high altitudes (mountains), it is very difficult to stop bleeding from a cut in the body.
R : At high altitude the atmospheric pressure is less than the blood pressure inside the body.
Sol. Answer (1)
A : is true
10. A : When liquid drops merge into each other to form a large drop, energy is released.
R : When liquid drops merge to form large drop surface tension decreases.
Sol. Answer (3)
A : is true
R : is false, because when large drop is formed, surface area gets reduced. Hence surface energy gets reduced
due to reduction in surface area not the surface tension.
11. A : Excess pressure inside a soap bubble is
r
T4
(symbols have their usual meanings).
R : The pressure difference across a curved surface of radius of curvature r is
r
T2
. There are two surfaces in a
soap bubble.
Sol. Answer (1)
A : is true
12. A : Buoyant force is always vertically upward.
R : Buoyant force is always opposite to the direction of acceleration due to gravity.
Sol. Answer (4)
A : is wrong, Buoyant force is not always vertically upwards
R : is wrong, Buoyant force is always opposite to the direction of effective acceleration.

13. A : Equation of continuity is A1v11 = A2v22 (symbols have their usual meanings).
R : Equation of continuity is valid only for incompressible liquids.
Sol. Answer (3)
A : is true
R : is false,
14. A : Atomizer is based on the principle of Bernoulli’s theorem.
R : Bernoulli’s theorem is based on the conservation of energy.
Sol. Answer (2)
A : is true
R : is true, but not the correct explanation
Correct explanation is, decrease in pressure forces the water to move up the tube and get sprayed.
15. A : The spiders and insects can run on the surface of water.
R : Buoyant force balances the weight of insects.
Sol. Answer (3)
A : true
R : is false, surface tension balances the weight of insect.

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