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 $\left(t_1/t_2\right)$ is:

• 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 $\frac{2-\sqrt{3}}{\sqrt{3}}$

Let height of tank $=H$

$\frac{H}{4}$ empty in $t_1$

$\frac{3H}{4}$ in $t_2$

Velocity of water going out through hole when high of water level is h,

$v=\sqrt{2gh}$

$\frac{dh}{dt}=\sqrt{2gh}$

$\int \frac{dh}{\sqrt{h}}=\sqrt{2g}\int dt$ ___(i)

${\int }_H^{3H/4}{h}^{-1/2}dh=\sqrt{2g}{\int }_0^{t_1}dt$

${\left[\frac{h^{-1/2+1}}{\frac{1}{2}}\right]}_H^{3H/4}=\sqrt{2g}{t}_1$

$2[{\left(\frac{3H}{4}\right)}^{1/2}-H^{1/2}]=\sqrt{2g}{t}_1$

$2\sqrt{H}(\sqrt{\frac{3}{4}}-1)=\sqrt{2g}{t}_1$ ___(ii)

From equation (1)

${\int }_{\frac{3h}{4}}^0\frac{dh}{\sqrt{h}}=\sqrt{2g}{\int }_0^{t_2}dt$

$\frac{h^{1/2}}{\frac{1}{2}}{|}_{3h/4}^0=\sqrt{2g}{t}_2$

$2(-{\left(\frac{3h}{4}\right)}^{\frac{1}{2}})=\sqrt{2g}{t}_2$

$-2\sqrt{h}\times \sqrt{\frac{3}{4}}=\sqrt{2g}{t}_2$ ____(iii)

(ii) $÷$ (iii)

$\frac{t_1}{t_2}=-\frac{\sqrt{\frac{3}{4}}-1}{\sqrt{\frac{3}{3}}}=-\frac{\sqrt{3}-2}{\sqrt{3}}$

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