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

Options

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

Answer

A

Explanation

Let `V` be the volume of the ball.
Net upward force = `V sigma g - Vrhog`
Net upward acceleration,
`a = (Vsigma g - Vrhog)/(Vrho) = ((sigma-rho)g)/(rho)`
velocity at the surface, = `sqrt((2(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-rho)g)/(grho) = ((sigma)/(rho)-1)h`.