Question

An open container of height H and Radius R is completely filled with water. A small orifice of radius r is made near the bottom of container. Time elapsed in emptying half of the container, is

Answer 1

Correct option is (D) \(\frac{R^2}{r^2}\sqrt{\frac{H}g}\)

We know that, time to empty a container if orifice at the bottom

t = \(\frac{A}{A_0}\sqrt{\frac{2H}{g}}\)

given \(\because H = H/2\)

t = \(\frac{\pi R^2}{\pi r^2}\sqrt{\frac{2H/2}g}\)

Where A₀ = area of orifice 

t = time

t = \(\frac{R^2}{r^2}\sqrt{\frac{H}g}\)