A body of density `rho` is dropped from a height h into a lake of density `sigma(sigmagtrho)`. The maximum depth the body sinks inside the liquid is (neglect viscous effect of liquid)

Options

A. `((hsigma)/(rho-sigma))`
B. `hsigma/(rho)`
C. `h((rho)/(rho-sigma))`
D. `h((rho)/sigma-1)`

Answer

A

Explanation

If `v_(0)-sqrt(2gh)` is the velocity downwards just when the body enters the liquid the final velocity at the maximum depth it will sink is zero. Its acceleration is `a`.
Hence `(-)y=` maximum depth
`(v_(0)^(2))/(2a)=(2gh)/(2g(rho/sigma-1))`
`:. Y=(hsigma)/(rho-sigma)`