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
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.
$$\frac{{Sh}}{V}g$$
Answer :
$$\frac{{Sh}}{V}g$$
Solution :
Let the density of water be $$\rho ,$$ then the force by escaping liquid on container $$ = \rho S{\left( {\sqrt {2gh} } \right)^2}$$
$$\therefore $$ Acceleration of container
$$\eqalign{ & a = \frac{{2\rho Sgh - \mu \rho Vg}}{{\rho V}} = \left( {\frac{{2Sh}}{V} - \mu } \right)g \cr & {\text{Now,}}\,\mu = \frac{{Sh}}{V} \cr & \therefore a = \frac{{Sh}}{V}g \cr} $$
Let the density of water be $$\rho ,$$ then the force by escaping liquid on container $$ = \rho S{\left( {\sqrt {2gh} } \right)^2}$$
$$\therefore $$ Acceleration of container
$$\eqalign{ & a = \frac{{2\rho Sgh - \mu \rho Vg}}{{\rho V}} = \left( {\frac{{2Sh}}{V} - \mu } \right)g \cr & {\text{Now,}}\,\mu = \frac{{Sh}}{V} \cr & \therefore a = \frac{{Sh}}{V}g \cr} $$
