Question

A container of height h is completely filled with water. The container is placed on a frictional surface with coefficient of friction and a small hole is punctured at a depth on container wall. If area of hole is a and area of base of container sA (a << A), then the value of for which the container remains stationary is (g = 10 m/s)

Answer 1

The coefficient of friction is a/A

Explanation:

The hole is at depth h/2

The velocity of water coming out from the hole

$v=\sqrt{2g(h/2)}$

Volume of liquid in container = Area of base × Height

$\implies V=Ah$

If the density of water is $\rho$

Then weight of the water in container = Volume × Density × g

$=Ah\rho g$

If coefficient of friction is $\mu$

Then the frictional force acting on the container

$F=\muAh\rho g$

The force on the container due to the water coming out of the hole

$F=\rho a v^2$

$\implies F=\rho agh$

For the container to be stationary

$\muAh\rho =\rho agh$

$\implies \mu=a/A$

Hope this answer is helpful.

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