Q 1 :

A spherical ball of radius 1×10-4 m and density 105 kg/m3 falls freely under gravity through a distance h before entering a tank of water. If after entering in water the velocity of the ball does not change, then the value of h is approximately

(The coefficient of viscosity of water is 9.8×10-6 Ns/m2)            [2024]

  • 2296 m

     

  • 2249 m

     

  • 2518 m

     

  • 2396 m

     

(3)   

          Terminal velocity, vT=2g9R2[ρB-ρL]η

         vT=29×9.8×(10-4)29.8×10-6[105-103]=220 m/s

          When ball fall from height (h), v=2gh

          Now h=v22g=(220)22×9.82518 m

 



Q 2 :

A small spherical ball of radius r, falling through a viscous medium of negligible density has terminal velocity 'v'. Another ball of the same mass but of radius 2r, falling through the same viscous medium will have terminal velocity                                  [2024]

  • v/2

     

  • v/4

     

  • 4v

     

  • 2v

     

(1)   

       Since density is negligible hence Buoyancy force will be negligible

        At terminal velocity,

        Mg=6πηrv

       v1r (as mass is constant)

        Now, vv'=r'r

        r'=2r

       So, v'=v2

 



Q 3 :

Small water droplets of radius 0.01 mm are formed in the upper atmosphere and falling with a terminal velocity of 10 cm/s. Due to condensation, if 8 such droplets are coalesced and formed a larger drop, the new terminal velocity will be _______ cm/s.                    [2024]



(40)            m=mass of small drop, M=mass of bigger drop

                 Vt=29r2(ρ-σ)gηVtr2

                 8m=M8r3=R3R=2r

                 Radius double so Vt becomes 4 times:

                  Vt=4×10=40cm/s

 



Q 4 :

A small ball of mass m and density ρ is dropped in a viscous liquid of density ρ0. After some time, the ball falls with constant velocity. The viscous force on the ball is:   [2024]

  • mg(ρ0ρ-1)

     

  • mg(1-ρρ0)

     

  • mg(1-ρρ0)

     

  • mg(1-ρ0ρ)

     

(4)

At constant velocity, Fv+FB=mg

FB=ρ0·(mρ)·g

Fv=mg-FB=mg-mρρ0gFv=mg(1-ρ0ρ)



Q 5 :

A small steel ball is dropped into a long cylinder containing glycerine. Which one of the following is the correct representation of the velocity-time graph for the transit of the ball?       [2024]

  •  

  •  

  •  

  •  

(4)

mg-FB-Fv=ma

(ρ43πr3)g-(ρL43πr3)g-6πηrv=mdvdt

Let 43mπR3g(ρ-ρL)=K1 and 6πηrm=K2

dvdt=K1-K2v  dv=(K1-K2v)dt

0vdvK1-K2v=0tdt

-1K2ln [K1-K2v]0v=t    ln(K1-K2vK1)=-K2t

K1-K2v=K1e-K2t    v=K1K2[1-e-K2t]



Q 6 :

A small rigid spherical ball of mass M is dropped in a long vertical tube containing glycerine. The velocity of the ball becomes constant after some time. If the density of glycerine is half of the density of the ball, then the viscous force acting on the ball will be (consider g as acceleration due to gravity)          [2025]

  • 32Mg

     

  • Mg2

     

  • Mg

     

  • 2 Mg

     

(2)

At terminal velocity a = 0

F=Vρ2g=Mg2

  f=MgMg2=Mg2



Q 7 :

Given below are two statements:

Statement-I : The hot water flows faster than cold water.

Statement-II : Soap water has higher surface tension as compared to fresh water.

In the light above statements, choose the correct answer from the options given below:           [2025]

  • Statement-I is false but Statement-II is true

     

  • Statement-I is true but Statement-II is false

     

  • Both Statement-I and Statement-II are true

     

  • Both Statement-I and Statement-II are false

     

(2)

Hot water flows faster because of less viscosity and soap water has less surface tension because bubbles are easily formed.



Q 8 :

A solid steel ball of diameter 3.6 mm acquired terminal velocity 2.45×102 m/s while falling under gravity through an oil of density 925 kg m3. Take density of steel as 7825 kg m3 and g as 9.8 m/s2. The viscosity of the oil in SI unit is          [2025]

  • 2.18

     

  • 2.38

     

  • 1.68

     

  • 1.99

     

(4)

vT  29(ρ0ρl)r2gη

η=29(78259252.45×102)×(1.8)2×106×9.8

η=1.99



Q 9 :

A small ball of mass M and density ρ is dropped in a viscous liquid of density ρ0. After some time, the ball falls with a constant velocity. What is the viscous force on the ball?           [2023]

  • F=Mg(1-ρ0ρ)

     

  • F=Mg(1+ρρ0)

     

  • F=Mg(1+ρρ0)

     

  • F=Mg(1+ρ0ρ)

     

(1)

For constant velocity Fnet=0

Fvis+ρ0vg=ρvg

Fvis=(ρ-ρ0)vg

=ρvg(1-ρ0ρ)=Mg(1-ρ0ρ)



Q 10 :

Eight equal drops of water are falling through air with a steady speed of 10 cm/s. If the drops coalesce, the new velocity is  [2023]

  • 10 cm/s

     

  • 40 cm/s

     

  • 16 cm/s

     

  • 5 cm/s

     

(2)

vr2

v1v2=(rR)2

8·43πr3=43πR3

R=2r

10v2=(12)2v2=40 cm/s



Q 11 :

Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R.

Assertion A: A spherical body of radius (5±0.1) mm having a particular density is falling through a liquid of constant density. The percentage error in the calculation of its terminal velocity is 4%.

Reason R: The terminal velocity of the spherical body falling through the liquid is inversely proportional to its radius.

In the light of the above statements, choose the correct answer from the options given below          [2023]

  • Both A and R are true but R is NOT the correct explanation of A

     

  • Both A and R are true and R is the correct explanation of A

     

  • A is false but R is true

     

  • A is true but R is false

     

(4)

Terminal velocity of a spherical body in liquid

Vtr2 ΔVtVt=2·Δrr

ΔVtVt×100%=20.15×100=4%

Also Vtr2

Reason R is false



Q 12 :

A spherical ball of radius 1 mm and density 10.5 g/cc is dropped in glycerine of coefficient of viscosity 9.8 poise and density 1.5 g/cc. The viscous force on the ball when it attains constant velocity is 3696×10-x N. The value of x is. (Given: g=9.8 m/s2 and π=227) ________ .           [2023]



(7)

When the ball attains terminal velocity

Fv=(mg-FB)  (a=0)

     =Vσbg-Vρg=V(σb-ρ)

     =43π(10-3)3×9.8(10.5-1.5)×103

      =3696×10-7 N

So, x=7



Q 13 :

A metal block of base area 0.20 m2 is placed on a table, as shown in the figure. A liquid film of thickness 0.25 mm is inserted between the block and the table. The block is pushed by a horizontal force of 0.1 N and moves with a constant speed. If the viscosity of the liquid is 5.0×10-3 poiseuille, the speed of the block is _______ ×10-3 m/s.        [2023]



(25)

|F|=ηAΔvΔh : 0.1=5×10-3×0.2×v0.25×10-3

v=0.025 m/s or v=25×10-3 m/s



Q 14 :

An air bubble of diameter 6 mm rises steadily through a solution of density 1750 kg/m3 at the rate of 0.35 cm/s. The coefficient of viscosity of the solution (neglect density of air) is _______ P as (given, g=10 ms-2).                    [2023]



(10)

Since the bubble is moving at constant speed, the force acting on it is zero.

B=Fv

43πR3ρg=6πηRv

η=2R2ρg9v=2×(3×10-3)2×1750×109×0.35×10-2=10 Pas



Q 15 :

Sixty four rain drops of radius 1 mm each falling down with a terminal velocity of 10 cm/s coalesce to form a bigger drop. The terminal velocity of the bigger drop is ______ cm/s.                [2026]



(160)

VT=2r2g9η(σ-ρ)

VTr2

64 drops

64(43πR13)=43πR23

R2=4R1

(VT)1(VT)2=(R1R2)2=(14)2

10(VT)2=116

(VT)2=160 cm/sec



Q 16 :

A small metallic sphere of diameter 2 mm and density 10.5 g/cm3 is dropped in glycerine having viscosity 10 Poise and density 1.5 g/cm3 respectively. The terminal velocity attained by the sphere is ____ cm/s. 

(π=227 and g=10 m/s2)       [2026]

  • 3.0

     

  • 1.0

     

  • 2.0

     

  • 1.5

     

(3)

VT=2r2g9η(ρb-ρ)

VT=29·(1)2×1010(10.5-1.5)

VT=2 cm/sec.



Q 17 :

A ball of radius r and density ρ dropped through a viscous liquid of density σ and viscosity η attains its terminal velocity at time t, given by t=Aρarbηcσd  where A is a constant and a,b,c and d are integers. The value of b+ca+d is _______.    [2026]



(1)

T=(ML-3)aLb(ML-1T-1)c(ML-3)d

T=Ma+c+dL-3a-c-3d+bT-c

on comparing

c=-1;  a+c+d=0;  -3a-c-3d+b=0

b=2;  a+d=1

b+c=1



Q 18 :

The terminal velocity of a metallic ball of radius 6 mm  in a viscous fluid is 20 cm/s. The terminal velocity of another ball of same material and having radius 3 mm, in the same fluid will be ______ cm/s.        [2026]



(5)

We know:

Terminal velocity(radius)2

(vT)1(vT)2=(63)2

(vT)2=(vT)14=5 cm/sec