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

     

(C)   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

     

(A)   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