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 $\frac{ h }{2}$ and from $\frac{ h }{2}$ to bottom is

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

Answer 1

Answer: (C)

According to the relation
$t=\frac{A}{A_{0}} \sqrt{\frac{2}{g} H}-\frac{A}{A_{0}} \sqrt{\frac{2 H'}{g}}$
$t =\frac{ A }{ A _{ o }} \sqrt{\frac{2}{ g }} \times\left(\sqrt{ H }-\sqrt{ H'}\right)$
$t \propto\left(\sqrt{ H }-\sqrt{ H'}\right)$
$\frac{ t _{1}}{ t _{2}}=\frac{\sqrt{ H }-\sqrt{\frac{ H }{2}}}{\sqrt{\frac{ H }{2}}-\sqrt{0}}=\frac{1-\frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}}}=\sqrt{2}-1$