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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 time `t_1`. Next, the ball is released and it falls through the same height before striking the surface of a liquid of density of `d_L` (a) If `dltd_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. (b) Is the motion of the ball simple harmonic? (c) If `d=d_L`, how does the speed of the ball depend on its depth inside the liquid? Neglect all frictional and other dissipative forces. Assume the depth of the liquid to be large. |
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Answer» Correct Answer - A::B::C::D In elastic collision with the surface, direction of velicity is reversed but its magnitude remains the same, Therefore, time of fall=time of rise, or, time of fall `=(t_(1))/(2)` Hence, velocity of the ball just before it collides with liquids is `v=g(t_(1))/(2)` Retardation inside the liquid `a=("upthrust-weight")/("mass")=(Vd_(L)g-Vdg)/(Vd)` `=((d_(L)-d)/(d))g` Time taken to come to rest under this retardation will be `t=v/a=(g t_(1))/(2a)=(g t_(1))/(2((d_(L)-d)/(d))g)` `=(dt_(1))/(2(d_(L)-d))` Same will be the time to come to back on the liquid surface. therefore, (a) `(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)` (b) The motion of the ball is periodic but not simple harminic because the acceleration of the ball is g in air and `((d_(L)-d)/(d)) g` inside the liqudi which is not proportional to the displacenent, which is necessary and sufficient condition for SHM. (c ) When `(d_(L)=d)`, retardation or acceleration inside teh liquid becones zero (upthrust-weight). Therefore, the ball will continue to move with constant velocity `v=(gt_(1))/(2)` inside the liquid. |
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