Q. 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 water. If $T_{1} = T_{2}$, then the value of $\eta$ is
A) 2
B) 3
C) 4
D) $2\sqrt{2}$
Explanation
(c) $t = \frac{A}{a}\sqrt{\frac{2}{g}}\left\lbrack \sqrt{H_{1}} - \sqrt{H_{2}}\ \right\rbrack$
Now, $T_{1} = \frac{A}{a}\sqrt{\frac{2}{g}}\left\lbrack \sqrt{H} - \sqrt{\frac{H}{\eta}} \right\rbrack$
and $T_{2} = \frac{A}{a}\ \sqrt{\frac{2}{g}}\ \left\lbrack \sqrt{\frac{H}{\eta}} - \sqrt{0} \right\rbrack$
According to problem $T_{1} = T_{2}$
$\therefore\sqrt{H} - \sqrt{\frac{H}{\eta}} = \sqrt{\frac{H}{\eta}} - 0 \Rightarrow \sqrt{H} = 2\sqrt{\frac{H}{\eta}} \Rightarrow \eta = 4$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.