Question

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. $\frac{x\pi R^{2}n_{1}}{n_{2}}$
• B. $\frac{x\pi R^{2}n_{2}}{n_{1}}$
• C. $\frac{2\pi Rn_{1}}{n_{2}}$
• D. $\pi R^{2}x$

Answer 1

Answer: (B)

$\frac{x\pi R^{2}n_{2}}{n_{1}}$

Answer 2

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. $\therefore \frac{n_{2}}{n_{1}} =\frac{\frac{d h}{d t}}{\frac{d x}{d t}}$ $\therefore\frac{d h}{d t} =\frac{n_{2}}{n_{1}} \times \frac{d x}{d t}$ or change in actual rate of flow $=\frac{n_{2}}{n_{1}} \times$ change in apparent rate of flow or $\frac{d h}{d t}=\frac{n_{2}}{n_{1}} \times x cm / min$ Multiplying both sides by $\pi R^{2}$, we have $\frac{d h}{d t} \times \pi R^{2}=\frac{n_{2}}{n_{1}} \times x \times \pi R^{2}$ $\therefore$ Amount of water drained $=x \pi R^{2} \frac{n_{2}}{n_{1}}$