A small ball of density `rho` is immersed in a liquid of density `sigma(sigma gt rho)` to a depth h and then released. The height above the surface of water up to which the ball will jump is

Options

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

Answer

A

Explanation

Let `V` be the volume of the ball.
Net upward acceleration.
`a=(Vsigmag-Vrhog)/(Vrho)=((sigma-rho)gh)/rho`
If `h_(a)` is the height in air to which the ball rises, then
`0-(2(sigma-rho)gh)/rho=2(-g)h_(a)`
`:. H_(a)=((sigma-r)gh)/(grho)=(sigma/rho-1)h`