A ball of density d is dropped on to a horizontal solid surface. It bounces elastically from the surface and returns to its original position in a time ` t_1` Next, the ball is released and its falls through the same height before striking the surface of a liquid of density ` d _ L`. Neglect all frictional and other dissipative force. Assume the depth of the liquid to be large.
If ` d lt d _ L ` obtain an expression (in terms of d, ` t_1 and d_L ` ) for the time ` t _ 2 `the ball takes to come back to the position from which it was released.

Options

A. ` 2t_1 `
B. ` (dt_1 )/((d_L- d )) `
C. ` (dt_1)/(2 (d_L- d )) `
D. ` (2dt_1 )/(( d _L - d )) `

Answer

C

Explanation

In elastic collision with the surface, direction of velocity is reversed but its magnitude remains the same. Therefore,
time of fall = time of rice. Or time of fall ` = (t_1 )/(2) `
Hence, velocity of the ball just before it collides with liquid is ` v = g (t _ 1 )/(2) " " ...(i) `
Retardation inside the liquid ` a = ("Upthrust - weight")/("mass") = (Vd_L g - V dg )/(Vd ) = ((d_L - d )/(d)) g " " `...(ii)
Time taken to come to rest under this retardation will be ` t = (V)/(a) = (g t _ 1 )/( 2 a ) = (g t _1)/(2((d_L- d )/(d ))g ) = (d t _ 1)/(2(d _ L - d )) `
same will be the time to come back on the liquid surface ` because ` a is constant. Therefore,
` t_2 ` = time the ball takes to came back to the position from where it was released
` = t _ 1 + 2t = t _ 1 + (dt _ 1 )/(d_L - d ) = t _ 1 [ 1 + (d)/(d_L - d )] or t_2 = (t_1 d _ L)/(d _ L - d ) `