Question

In the figure shown a light container is kept on a horizontal rough surface of coefficient of friction $\mu =\frac{Sh}{V}$. 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

• A. $\frac{V}{Sh}g$
• B. $g$
• C. $zero$
• D. $\frac{Sgh}{V}$

Answer 1

The correct option is D $\frac{Sgh}{V}$

Given,

Friction coefficient, $\mu =\frac{Sh}{V}$

Area of hole is $S$

Velocity of liquid, $v=\sqrt{2gh}$

Force, $F=\frac{dP}{dt}=rateofflow\times velocity=\left(\frac{\rho Svdt}{dt}\right)v=\rho S{v}^2=\rho S\left(2gh\right)$

Friction force, $F_r=\mu Mg=\frac{Sh}{V}\left(\rho V\right)g=\rho shg$

Net force = Fluid force – friction force

$F_{net}=2\rho Sgh-\rho Sgh=\rho Sgh$

Acceleration of container, $a^{\prime }=\frac{F_{net}}{M}=\frac{\rho Sgh}{\rho V}=\frac{Sgh}{V}$

heance acceleration of container is $\frac{Sgh}{V}$.