A wooden ball of density ` rho` is immersed in a liquid of density `sigma` to a depth H and then released. The height h above the surface of which the ball rises will be

Options

A. H
B. `(sigma)/(rho)`
C. `((sigma-rho)/(rho))H`
D. `(rho)/(sigma)H`

Answer

C

Explanation

Let the volume of wooden ball be V. Weight of the ball `=V rho g`. Upward thrust on the ball due to the liquid `=V sigma g`. Net upward force on the ball `=V sigma g- V rho g= V g(sigma-rho)`
Upward ac celeration in the ball
`a=("force")/(mass)=(V g(sigma-rho))/(V rho)=((sigma-rho)g)/(rho)`
Velocity on reaching the surface `v=sqrt(2 a H)` , If h is the height to which the ball rises above the surface of liquid, then `v=sqrt(2 gh)` . Hence
`sqrt(2 a H)=sqrt(2 gh)` or `2a H=2 gh`
or `h=(a H)/(g)=((sigma-rho)/(rho))(g H)/(g)= ((sigma-rho)/(rho))H`