A body of density `rho` is dropped from rest from height h (from the surface of water) into a lake of density of water `sigma(sigmagt rho)`. Neglecting all disspative effects, the acceleration of body while it is in the take is

Options

A. `(sigmag)/(rho)` upwards
B. `(sigmag)/(rho)` downwards
C. `g(sigma/rho-1)`upwards
D. `g(rho/sigma-1)` upwards

Answer

D

Explanation

The upward buoyant force `B=rhogV, ` where `V` is the volume of the body.
Hence, the equation of motion is
`B-mg=ma,` where `a` is the upward acceleration i.e.
`rhogV-sigmaVg=sigmaV_(a)`
Hence `a=(g(rho-sigma))/(sigma)=g((rho)/(sigma)-10)` upward