An object of density `sigma(ltrho)` is dropped from rest at a height h into a liquid of density `rho` kept in a tall vertical cylindrical container. Neglect all dissipative effects and assume that there is no spilling of the liquid.
The time taken by the object to reach the starting point (from which it is dropped) is

Options

A. `2sqrt((2h)/g)`
B. `4sqrt((2h)/g)`
C. `3sqrt((2h)/g)`
D. it will never reach starting point

Answer

B

Explanation

Time taken to reach the surface of liquid `t_(1)=sqrt((2h)/g`
Time taken by object to travel from surface of liqid to maximum depth
`t_(2)=sqrt((2h((sigma)/(rho-sigma)))/(g((rho-sigma)/sigma))`
As `rho=2sigma`
`t_(2)=sqrt((2h)/g)`
total time taken to reach startiing point is
`t=2t_(1)+2t_(2)=4sqrt((2h)/g)`