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 time taken for the level of water to fall from $h$ to $\frac{\text{h}}{2}$ and from $\frac{\text{h}}{2}$ to zero is

• A. $\sqrt{2}$
• B. $\frac{1}{\sqrt{2}}$
• C. $\sqrt{2} - 1$
• D. $\frac{1}{\sqrt{2} - 1}$

Answer 1

Answer: (C)

Time taken for the level to fall from H to H'
$\text{t} = \frac{\text{A}}{\text{A}_{0}} \sqrt{\frac{\text{2}}{\text{g}}} \left[\sqrt{\text{H}} - \sqrt{\text{H}^{′}}\right]$
According to problem the time taken for the level to fall from h to $\frac{\text{h}}{2}$
$\text{t}_{1} = \frac{\text{A}}{\text{A}_{0}} \sqrt{\frac{2}{\text{g}}} \left[\right. \sqrt{\text{h}} - \sqrt{\frac{\text{h}}{2}} \left]\right.$
and similarly time taken for the level to fall from $\frac{\text{h}}{2}$
$\text{t}_{2} = \frac{\text{A}}{\text{A}_{0}} \sqrt{\frac{2}{\text{g}}} \left[\right. \sqrt{\frac{\text{h}}{2}} - 0 \left]\right.$
$∴ \frac{\text{t}_{1}}{\text{t}_{2}} = \frac{1 - \frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}} - 0} = \sqrt{2} - 1$