Q. The apparent depth of water in cylindrical water tank of diameter $2R\ cm$ is reducing at the rate of $x\ cm/$minute when water is being drained out at a constant rate. The amount of water drained in $c.c.$ per minute is $(n_{1} =$ refractive index of air, $n_{2} =$ refractive index of water)
A) $x\ \pi\ R^{2}\ n_{1}/n_{2}$
B) $x\ \pi\ R^{2}\ n_{2}/n_{1}$
C) $2\ \pi\ R\ n_{1}/n_{2}$
D) $\pi\ R^{2}\ x$
Explanation
(b) Apparent depth $h^{'} = \frac{h}{\ _{air}\mu_{liquid}}$
$\Rightarrow \frac{dh^{'}}{dt} = \frac{1}{\ _{a}\mu_{\omega}}\frac{dh}{dt} \Rightarrow x = \frac{1}{\ _{a}\mu_{\omega}}\frac{dh}{dt}\ \ \ \ \left\lbrack \because\frac{dh^{'}}{dt} = x \right\rbrack$
$\Rightarrow \frac{dh}{dt} = \ _{a}\mu_{\omega}x$
Now volume of water $V = \pi R^{2}h$
$\Rightarrow \frac{dV}{dt} = \pi R^{2}\frac{dh}{dt} = \pi R^{2} \bullet \ _{a}\mu_{\omega}x$
$= \ _{a}\mu_{\omega}\pi R^{2}x = \frac{\mu_{\omega}}{\mu_{a}}\pi R^{2}x = \left( \frac{n_{2}}{n_{1}} \right)\pi R^{2}x$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.