Question

A cylinder of radius R has a hole of radius r at a depth H below the level of liquid in cylinder. At t = 0, the hole is opened. What is time taken to reduce level of liquid to H/2 ?

Answer 1

As liquid emerges out of hole O; the level of liquid above O in cylinder decreases. The velocity of efflux and rate of flow of liquid in cylinder decreases as time increases. Ley y be the instantaneous height of liquid above O. The instantaneous rate of flow of liquid

Let y+dy be level of liquid in cylinder at t = t + dt.

Then dV = πR² (-dy) . Therefore

Integrating Eqn. (1) within limits y = H to y = H/2 gives the desired time t. Therefore