Question

The apparent depth of water in cylindrical water tank of diameter $2Rcm$ is reducing at the rate of $xcm/$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. $\frac{x\pi R^2{n}_1}{n_2}$
• B. $\frac{x\pi R^2{n}_2}{n_1}$
• C. $\frac{2\pi R{n}_1}{n_2}$
• D. $\pi R^{2}x$

Answer 1

Answer: (B)

Let actual height of water of the tank be $h$, then

${}_1{n}_2=\frac{\text{ actual depth }}{\text{ apparent depth }}$
Also ${}_1{n}_2=\frac{n_2}{n_1}$
$\therefore \frac{n_2}{n_1}=\frac{h}{x}$
where $x$ is a apparent depth.
$\therefore \frac{n_2}{n_1}=\frac{\frac{dh}{dt}}{\frac{dx}{dt}}$
$\therefore \frac{dh}{dt}=\frac{n_2}{n_1}\times \frac{dx}{dt}$
or change in actual rate of flow $=\frac{n_2}{n_1}\times$ change in apparent rate of flow or $\frac{dh}{dt}=\frac{n_2}{n_1}\times xcm/min$
Multiplying both sides by $\pi R^2$, we have
$\frac{dh}{dt}\times \pi R^2=\frac{n_2}{n_1}\times x\times \pi R^2$
$\therefore$ Amount of water drained $=x\pi R^2\frac{n_2}{n_1}$