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 fourths of the tank is emptied is $t_2$ seconds, then ratio $(t_1/t_2)$ is

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

Answer 1

Answer: (D)

$\frac{2 - \sqrt{3}}{\sqrt{3}}$

Answer 2

Let h be the total height of the water in the tank above the hole. Let A be the area of the cross-section of the tank and a be the area of the cross-section of the hole.
$\therefore \, -A \frac{dh}{dt} = a \sqrt{2gh}$ or $dt = -\frac{A}{a\sqrt{2g}} \frac{dh}{\sqrt{h}}$
$\therefore$ Required ratio $\frac{t_1}{t_2} = \frac{\int\limits_h^{3h/4} \frac{dh}{\sqrt{h}}}{\int\limits_{3h/4}^0 \frac{dh}{\sqrt{h}}}$
$\frac{t_{1}}{t_{2}}= \frac{\left[\sqrt{\frac{3h}{4}} -\sqrt{h}\right]}{\left[0 - \sqrt{\frac{3h}{4}}\right]} = \frac{2 - \sqrt{3}}{\sqrt{3}}$