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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 timet_1 Next, the ball is released and its falls through the same height before striking the surface of a liquid of densityd _ L. Neglect all frictional and other dissipative force. Assume the depth of the liquid to be large. Ifd lt d _ Lobtain an expression (in terms of d,t_1andd_L) for the timet _ 2 the ball takes to come back to the position from which it was released. |
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Answer» `2t_1 ` timeoffall= time ofrice. Ortime offall `= (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 )] ort_2= (t_1d _ L)/(d _ L - d ) ` |
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