1.

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 :

Answer»

`(d_(1))/(d_(2))sqrt((2H)/G)`
`(d_(2))/(d_(1))sqrt((2h)/g)`
`(d_(1))/(d_(2)-d_(1))sqrt((2h)/g)`
`(d_(2)-d_(1))/(d_(2))sqrt((2h)/g)`

Solution :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).


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