Question

A large tank filled with water to a height ‘h’ is to be emptied through a small hole at the bottom. The ratio of times taken for the level of water to fall from h to $\frac{h}{2}$ and from $\frac{h}{2}$ to zero is

• A.
• B.
• C.
• D.

Answer 1

The correct option is C

Time taken for the level to fall from H to H' t = $\frac{A}{A_0}\sqrt{\frac{2}{g}}[\sqrt{H}-\sqrt{H^{\prime }}]$ According to problem- the time taken for the level to fall from h to $\frac{h}{2}{t}_1=\frac{A}{A_0}\sqrt{\frac{2}{g}}[\sqrt{h}-\sqrt{\frac{h}{2}}]$
and similarly time taken for the level to fall from $\frac{h}{2}tozero{t}_2=\frac{A}{A_0}\sqrt{\frac{2}{g}}[\sqrt{\frac{h}{2}}-0]$
$\therefore \frac{t_1}{t_2}=\frac{1-\frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}}-0}=\sqrt{2}-1$.