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

Answer

B

Explanation

`2a_(1)S_(1)=2a_(2)S_(2)`
Here `a_(1)`=acceleration inside the liquid
`=("upthrust-weight")/("mass")=(Vsigma g- V rho g)/(rho)=((sigma)/(rho)-1)g`
`a_(2)`= retardation in air=g
`therefore " " S_(2)=(a_(1))/(a_(2)). S_(1)=((sigma)/(rho)-1)h`