Question

A tank full of water has a small hole at its bottom. If one-fourth of the tank is emptied in $t_1$ seconds and the remaining three fourth of the tank is emptied in $t_2$ seconds, then ratio $\left(t_1/t_2\right)$

• A. $\sqrt{3}$
• B. $\sqrt{2}$
• C. $\frac{2-\sqrt{2}}{\sqrt{2}}$
• D. $\frac{2-\sqrt{3}}{\sqrt{3}}$

Answer 1

The correct option is D $\frac{2-\sqrt{3}}{\sqrt{3}}$

General equation for time required to empty the tank from water level H to h is given as :-

$T=\frac{2A}{a\sqrt{2g}}[\sqrt{H}-\sqrt{h}]$

Where, A and a are c/s area of tank and opening in the tank

H = Initial liquid level in tank (H)

h = final water level in tank.

So,

$t_1=\frac{2A}{a\sqrt{2g}}[\sqrt{H}-\sqrt{\frac{3H}{4}}]$

$=\frac{2A}{a\sqrt{2g}}\sqrt{H}\left(\frac{2-\sqrt{3}}{2}\right)$

Similarly,

$t_2=\frac{2A}{a\sqrt{2g}}[\sqrt{\frac{3H}{4}}-0]$

$=\frac{2A}{a\sqrt{2g}}\left(\frac{\sqrt{3H}}{2}\right)$

So, $\frac{t_1}{t_2}=\frac{2-\sqrt{3}}{\sqrt{3}}$