Question

A tank full of water has a small hole at the bottom. If one-fourth of the tank is emptied in $t_1$ seconds and the remaining three-fourths of the tank is emptied in $t_2$ seconds. Then the ratio $\frac{t_1}{t_2}$ is

• A. $\sqrt{3}$
• B. $\sqrt{2}$
• C. $\frac{1}{\sqrt{2}}$
• D. $\frac{2}{\sqrt{3}}-1$

Answer 1

Answer: (D)

$\frac{2}{\sqrt{3}}-1$

time for tank empty to $\frac{h}{4}=t_1$

we know that

$v=u-gt$

$V=\sqrt{2g\left(\frac{3h}{4}\right)};u=\sqrt{2gh}$

$t_1=\frac{\sqrt{2gh}-\sqrt{2g(\frac{3h}{4}})}{g}$

$t_2=\sqrt{2g\left(\frac{3h}{4}\right)}$

$\frac{t_1}{t_2}=-\left(\frac{\frac{\sqrt{3}}{2}-1}{\frac{\sqrt{3}}{2}}\right)=\frac{2-\sqrt{3}}{\sqrt{3}}$

Correct answer is $\frac{2}{\sqrt{3}}-1$