Newton's Law of Viscosity DPP-16

1. The relative velocity of two consecutive layers is 8 cm/s. It the perpendicular distance
between the layers is 0.1 cm, then the velocity gradient will be–
(1) 8 sec–1 (2) 80 sec–1 (3) 0.8 sec–1 (4) 0.08 sec–1

2. A plate of area 100 cm2 is placed on the upper surface of castor oil, 2 mm thick. Taking
the coefficient of viscosity 10 poise, the horizontal force necessary to move the plate with
a uniform velocity of 2 cm/s is
(1) 1 N (2) 2 N (3) 0.1 N (4) 0.2 N

3. There is a 1 mm thick layer of glycerine between a plate of area 100 cm2 and a large plate.
If the coefficient of viscosity of glycerine is 1.0 kg/ms, then what force is required to move
the smaller plate with a velocity of 7 cm/s.
(1) 0.75 (2) 0.07 (3) 0.7 (4) 7

4. There is a 3mm thick layer of glycerine between a plate of area 10–3 m2 and a large plate.
If the coefficient of viscosity of glycerine is 2 kg/ms, then what force is required to move
the smaller plate with a velocity of 12 cm/s.
(1) 8N (2) 0.8N (3) 0.08N (4) 1.6N

5. The velocity of water in a river is 15 m/s near the surface. If the river is 5m deep, find the
shearing stress between the horizontal layers of water. The coefficient of viscosity of
water = 10–2 poise.
(1) 3 × 10–2 N/m3 (2) 0.33 × 10–2 N/m3
(3) 3.3 × 10–2 N/m3 (4) 3 × 10–3 N/m3

6. A cubical block of side 'a' and density '𝛒' slides over a fixed inclined plane with constant
velocity 'v'. There is a thin film of viscous fluid of thickness 't' between the plane and the
block. Then find the coefficient of viscosity of the thin film.

37°
3ρagt 4ρagt 5ρagt 5ρagt
(1) (2) (3) (4)
5v 5v 3v 4v
7. The velocity of water in a river is 18 km/h at the surface. If the river is 5 m deep and the
flow is streamlined, find the shearing stress between the horizontal layers of water
assuming uniform velocity gradient. Viscosity of water is 10–3 Poiseuille.
(1) 1 × 10–4 N/m2 (2) 1 × 10–3 N/m2
(3) 3 × 10–2 N/m2 (4) 2 × 10–4 N/m2

8. Velocity of water in a river is

(1) Same everywhere
(2) More in the middle and less near its banks
(3) Less in the middle and more near its banks
(4) Increase from one bank to other bank

9. A viscous fluid is flowing through a cylindrical tube. The velocity of the liquid in contact
with the walls of the tube is
(1) Zero (2) Maximum
(3) In between zero and maximum (4) Equal to critical velocity

10. A viscous fluid is flowing through a cylindrical tube. The velocity distribution of the fluid
is best represented by the diagram

(1) (2)

(3) (4) None of these

 Answer Key
Question 1 2 3 4 5 6 7 8 9 10
Answer 2 3 3 3 2 1 2 2 1 3

SOLUTIONS
1. (2)
dv 8
= = 80 per sec.
dx 0.1

2. (3)
ηAν 1×100×10−4 ×2×10−2
F= = = 0.1 N
ℓ 2×10−3

3. (3)
Δv 1.0×100×10−4 ×(7×10−2 )
Required force F = A = = 0.7 N
Δx 10−3

4. (3)
Δv 2×10−3 ×12×10−2
F = ηA = = 0.08N
Δy 3×10−3

5. (2)
Δv 15
Velocity gradient = = 3s −1
Δy 5
F Δv
shearing stress = =η ⇒ 3 × 10−3 N/m2
A Δy

6. (1)
dv
F = ηA
dy
dv
mg sin 3 7 º = ηA
dy
3 v
⇒ (a3 ρ)g × = η. a2 .
5 t
3ρagt
⇒η=
5v

7. (2)
As velocity at the bottom of the river will be zero,
dy 18×103
Velocity gradient = = 1s −1
dy 60×60×5
F dv
Shear stress = =η = 10 × 1 = 1 × 10–3 N/m2.
–3
A dy

8. (2)
More in the middle and less near its banks

9. (1)
Zero

10. (3)
 Stokes' Law and Terminal Velocity DPP-17

1. A spherical ball is dropped in a long column of viscous liquid. The speed v of the ball
varies as function of time as–

v
v v
(1) v (2) (3) (4)

(0,0) (0,0) t (0,0) t

(0,0) t t

2. If a ball is thrown in viscous liquid with a speed greater than terminal speed in same
direction of terminal speed than–
(1) Its speed increases upto infinite
(2) Its speed remains constant
(3) Its speed decreases and become constant
(4) Its speed decreases and become zero

3. A spherical body of diameter 'D' is falling in viscous medium. Its terminal velocity vt is
proportional to :-
(1) vt  D1/2 (2) vt  D3/2 (3) vt  D2 (4) vt  D5/2

4. Two drops of water which are falling in air are having mass ratio 1 : 27, what will be ratio
of their terminal speed–
(1) 1 : 9 (2) 1 : 4 (3) 1 : 3 (4) 3 : 1

5. An oil drop falls through air with a terminal velocity of 2 × 10–4 m/s. Viscosity of air is
𝐍−𝐬
10–5 and density of oil is 900 kg/m3 and g = 10 m/s2. Neglect the density of air as
𝐦𝟐
compared to that of oil, the radius of the drop is–
(1) 10–3 m (2) 10–4 m (3) 10–5 m (4) 10–6 m

6. An iron sphere is dropped into a viscous liquid. Which of the following represents its
acceleration (a) versus time (t) graph?
a a a
a

(1) (2) (3) (4)

O O t t
t t O O
7. Two equal drops are falling through air with a steady velocity of 5 cm/second. If two
drops coalesce to form one drop then new terminal velocity will be–
(1) 5 × (4)1/3 cm/s (2) 5√2 cm/s
5
(3) cm/s (4) 10 cm/s
√2

8. In a viscous medium terminal velocity (v) of a spherical body depends on its radius (r) as–
(1) v  r1/2 (2) v  r3/2 (3) v  r (4) v  r2

9. A small lead ball is falling freely in a viscous liquid. The velocity of the ball
(1) Goes on increasing (2) Goes on decreasing
(3) Remains constant (4) First increases and then becomes constant

10. A spherical body is moving in a viscous medium. Viscous force on the body does not
depend on
(1) Radius of the body (2) Density of the body
(3) Velocity of the body (4) Viscosity of the medium

11. A small spherical body is released from rest in a viscous medium. Its acceleration (a) varies
with velocity (v) as

a a
(1) (2)
a0
v v

a a
(3) (4)

v vT v

12. 27 identical drops of water are falling down vertically in air each with a terminal velocity
0.15 ms–1. If they combine to form a single bigger drop. What will be its terminal velocity?
(1) 0.3 ms–1 (2) 1.35 ms–1 (3) 0.45 ms–1 (4) zero

13. Estimate the speed of vertically falling raindrops from the following data. Radius of the
drops = 0.02 cm viscosity of air = 1.8 × 10–4 poise g = 10 m/s2 and density of water
103 kg/m3 neglect density of air :-
(1) 10 m/s (2) 3m/s (3) 5 m/s (4) None of these
14. A metal sphere of radius 1 mm and mass 50 mg falls vertically in glycerine. Find the
viscous force exerted by the glycerine on the sphere when the speed of the sphere is
1 cm/s. Density of glycerine = 1260 kg/m3 and coefficient of viscosity at room
temperature = 8.0 poise :-
(1) 3 × 10–4 N (2) 1.5 × 10–4 N (3) 6 × 10–6 N (4) 1.2 × 10–3 N

15. Spherical balls of radius 'r' are falling in a viscous fluid of viscosity '' with a velocity 'v'.
The retarding viscous force acting on the spherical ball is
(1) Inversely proportional to 'r' but directly proportional to velocity 'v'
(2) Directly proportional to both radius 'r' and velocity 'v'
(3) Inversely proportional to both radius 'r' and velocity 'v'
(4) Directly proportional to 'r' but inversely proportional to 'v'

16. A small drop of water falls from rest through a large height h in air; the final velocity is
(1) ∝ √h (2) ∝ h
(3) ∝ (1/h) (4) Almost independent of h

17. Eight drops of water, each of radius 2 mm are falling through air at a terminal velocity of
8 cm s–1. If they coalesce to form a single drop, then the terminal velocity of combined
drop will be :-
(1) 32 cm s–1 (2) 30 cm s–1 (3) 28 cm s–1 (4) 24 cm s–1

18. A small ball is left in a viscous liquid from very much height. Correct graph of its velocity
with time after it enters in liquid is :

B
A
velocity

C
D
time
(1) A (2) B (3) C (4) D
 Answer Key
Question 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
Answer 2 3 3 1 4 4 1 4 4 2 3 2 3 2 2
Question 16 17 18
Answer 4 1 3

SOLUTIONS
1. (2)

2. (3)

3. (3)
2r2 g
v= (ρs − ρ2 ) ⇒ v ∝ r 2  v  D2
9η

4. (1)
1 1
r1 m1 3 1 3 1
=[ ] =[ ] =
r2 m2 27 3
∵ V ∝ r2
VT1 r21 1 2 1
= =[ ] =
VT2 r22 3 9

5. (4)
2r2 (d−ρ)g
νT =
9η
2r2 ×900×10
2×10–4 = ; r = 10–6 m
9×10−5

6. (4)

7. (1)
4 4
2× πr 3 = πR3
3 3
R = 21/3 r
Now, VT  r2
VT = 22/3×5 = 41/3 × 5

8. (4)
2r2 g
v= (ρs − ρ2 ) ⇒ v ∝ r 2
9η

9. (4)
First increases then attains terminal velocity which is constant.
10. (2)
FV = 6πηrv
 FV º (independent of density)

11. (3)
(W−Th)−6πηrv
a=
m
Y = C – KX (straight line)

12. (2)
R = n1/3 r
v  r2
v1 r2 1
= =
v2 R2 9
v = 9 × 0.15 = 1.35 m/sec.

13. (3)
2 (σ−ρ)r2 g
Terminal velocity =
9 η
Substituting the values
VT = 5m/s

14. (2)
F = 6rv
F = 6 × 3.14 × 8 × 10–1 × 1 × 10–3 × 10–2
= 1.5 × 10–4 N

15. (2)
F = 6rv

16. (4)
F = 6rv

17. (1)
Let the radius of bigger drop is R and smaller drop is r then
4 4
πR3 = 8 × × πr 3
3 3
or R = 2r .....(i)
Terminal velocity, v  r2
v′ R2 2r 2
 = =( ) =4 (Using (i))
v r2 r
–1
or v' = 4v = 4 × 8 = 32 cm s

18. (3)
 Reynold's Number and Poiseuille's Equation DPP-18

1. If  = density, v = velocity, D = diameter and  = coefficient of viscosity. Reynold's number

can be given by
ρv ρvD ρη ηvD
(1) (2) (3) (4)
ηD η vD ρ

2. The Reynolds number of a flow is the ratio of

(1) Gravity to viscous force (2) Gravity force to pressure force
(3) Inertia forces to viscous force (4) Viscous forces to pressure forces

3. Viscosity of gases–
(1) Increases by increasing temperature (2) Increases by decreasing temperature
(3) Increases by decreasing pressure (4) Increases by increasing pressure

4. The cause of viscosity in gases is–

(1) cohesive force (2) adhesive force
(3) diffusion (4) conductivity

5. As the temperature of water increases, its viscosity

(1) Remains unchanged
(2) Decreases
(3) Increases
(4) Increases or decreases depending on the external pressure

6. Water flows in a streamlined manner through a capillary tube of radius a, the pressure
difference being P and the rate of flow Q. If the radius is reduced to a/2 and the pressure
increased to 2P, the rate of flow becomes
Q Q
(1) 4Q (2) Q (3) (4)
4 8

7. When a viscous liquid flows at a rate Q through a tube of radius r placed horizontally, a
pressure difference P develops across the ends of the tube. If the radius of the tube is
doubled and rate of flow also doubled. Then find out the pressure difference across the
ends of the tube.
P P
(1) (2) 8P (3) (4) 16P
8 16
 Answer Key
Question 1 2 3 4 5 6 7
Answer 2 3 1 1 2 4 1

SOLUTIONS
1. (2)

2. (3)

3. (1)

4. (1)

5. (2)

6. (4)
πpr4
Q= Q ∝ Pr 4 ( and l are constants)
8ηl
Q2 P2 r2 4 1 4 1 Q
 =( )( ) =2×( ) =  Q2 =
Q1 P1 r1 2 8 8

7. (1)
πPr4 Q1 P1 r41
Q= ⇒ =
8ηℓ Q2 P1 r42
Q pr4
⇒ =
2Q P2 (2r)4
P
⇒ P2 =
8