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Question

If d<dL obtain an expression (in terms of d,t1 and dL) for the time t2 the ball takes to come back to the position from which it was released.

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Solution

In elastic collision with the surface, direction of velocity is reversed but its magnitude remains the same.
Therefore. time of fall=time of rise.
or time of fall=t1t2
Hence, velocity of the ball just before it collides with liquid is
v=gt12 (i)
Retardation inside the liquid
a=upthrustweightmass
=VdLgVdgVd
=(dLdd)g (ii)
Time taken to come to rest under this retardation will be
t=va
=gt12a
=gt12(dLdd)g
=dt12(dLd)
Same will be the time to come back on the liquid surface.
Therefore,
t2=time the ball takes to came back to the position from where it was released
=t1+2t
=t1+dt1dLd
=t1[1+ddLd]
or t2=t1dLdLd

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