In the figure shown, a light container is kept on a horizontal rough surface of coefficient of friction `mu=(Sh)/V`. A very small hole of area S is made at depth 'h'. Water of volume 'V' is filled in the container. The friction is not sufficient to keep the container at rest. The acceleration of the container initially is

Options

A. `V/(Sh)g`
B. g
C. zero
D. `(Sh)/Vg`

Explanation

Let the density of water be `rho`, then the force by escaping liquid on container `=rhoS(sqrt(2gh))^(2)`
`:.` acceleration of container `a=(2rhoSgh-mu rhoVg)/(rhoV) =((2Sh)/V-mu)g`
Now `mu =(Sh)/V :. A=(Sh)/V g`