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$