Question

A large tank is filled with water to a height $H$. A small hole is made at the base of the tank. It takes $T_1$ time to decrease the height of water to $\frac{H}{\eta }(\eta >1),$ and it takes $T_2$ time to take out the rest of the water. If $T_1=T_2,$ then the value of $\eta$ is

• A. $2$
• B. $3$
• C. $4$
• D. $5$

Answer 1

The correct option is C $4$

We know that, time taken for the water to empty from height $H_1$ to $H_2$ which leaks from a container is given by -
$T=\frac{A}{a}\sqrt{\frac{2}{g}}\times (\sqrt{H_1}-\sqrt{H_2})$


Now, time taken to decrease height from $H$ to $\frac{H}{\eta }$ is
$T_1=\frac{A}{a}\sqrt{\frac{2}{g}}\times (\sqrt{H}-\sqrt{\frac{H}{\eta }})$ ........(1)
and,
Time taken to decrease height from $\frac{H}{\eta }$ to $0$,
$T_2=\frac{A}{a}\sqrt{\frac{2}{g}}\times (\sqrt{\frac{H}{\eta }}-\sqrt{0})$ ........(2)
According to question,
$T_1=T_2$
$\therefore \sqrt{H}-\sqrt{\frac{H}{\eta }}=\sqrt{\frac{H}{\eta }}-0$
$[$ from (1) and (2) $]$
$\Rightarrow \sqrt{H}=2\sqrt{\frac{H}{\eta }}$
$\Rightarrow \eta =4$