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