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

Options

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

Answer

C

Explanation

`a=((sigma-rho)/(rho))g,upsilon=sqrt((2(sigma-rho)gh)/(rho))`
`:. H=(upsilon^(2))/(2g)=((sigma-rho)/(rho))h`