A ball of material of specific gravity `d_(1)` falls from a height 'h' on the surface of a liquid of relative density `d_(2)` such that `d_(2) > d_(1)`. The time for which the body will be falling into the liquid is :

Options

A. `(d_(1))/(d_(2))sqrt((2h)/g)`
B. `(d_(2))/(d_(1))sqrt((2h)/g)`
C. `(d_(1))/(d_(2)-d_(1))sqrt((2h)/g)`
D. `(d_(2)-d_(1))/(d_(2))sqrt((2h)/g)`

Answer

C

Explanation

Velocity of fall of the body just before striking
`v=sqrt(2gh)`
Retarding force F = V `(d_(2)- d_(1)) g`
`therefore` Retardation a =`F/m=(V(d_(2)-d_(1))/(Vd_(1))g`
`=(d_(2)-d_(1))/(d_(1))g`
Time of fall `t=v/a=(sqrt(2gh))/((d_(2)-d_(1))/(d_(1))g)`
`=(d_(1))/(d_(2)-d_(1))xxsqrt((2h)/g)`
`therefore` Correct choice is (c).