Question

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

• A. 2
• B. 3
• C. 4
• D. $2\sqrt{2}$

Answer 1

Answer: (C)

$t=\frac{A}{a}\sqrt{\frac{2}{g}}\left[\sqrt{H_1}-\sqrt{H_2}\right]$
Now, $T_1=\frac{A}{a}\sqrt{\frac{2}{g}}\left[\sqrt{H}-\sqrt{\frac{H}{\eta }}\right]$
and $T_2=\frac{A}{a}\sqrt{\frac{T}{g}}\left[\sqrt{\frac{H}{\eta }}-\sqrt{0}\right]$
According to problem, $T_1=T_2$
$\therefore \sqrt{H}-\sqrt{\frac{H}{N}}-0$
$\Rightarrow \sqrt{H}=2\sqrt{\frac{H}{n}}$
$\Rightarrow n=4$