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 71 time to decrease the height of water to $\frac{H}{\eta }(\eta >1)$ and $T_2$ time to take out the rest of 11 water. If $T_1=T_2$ then the value of $\eta$ is

• A. 2
• B. 4
• C. 6
• D. 8

Answer 1

Answer: (B)

4

$t=\frac{A}{a}\sqrt{\frac{2}{g}}[\sqrt{H_1}-\sqrt{H_2}]$
$T_1=\frac{A}{a}\sqrt{\frac{2}{g}}\left[\sqrt{H_1}\sqrt{\frac{H}{\eta }}\right]$
$T_2=\frac{A}{a}\sqrt{\frac{2}{g}}\left[\sqrt{\frac{H}{\eta }-0}\right]$
Given, $T_1=T_2$
$\sqrt{H}-\sqrt{\frac{H}{\eta }}=\sqrt{\frac{H}{\eta }}-0$
$\Rightarrow \sqrt{H}=2\sqrt{\frac{H}{\eta }}$
$\Rightarrow \eta =4$