Question

The apparent depth of water in cylindrical water tank of diameter $2R$ cm 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. $\pi R^{2}x$
• D. $\frac{2\pi R{n}_1}{n_2}$

Answer 1

Answer: (B)

Let the apparent depth be $=d_A$ and real depth be $d_R$

$\frac{d_R}{d_A}=\frac{n_2}{n_1}$

$d_A=\frac{n_1}{n_2}{d}_R$

$\Delta d_A=\frac{n_1}{n_2}\Delta d_R$

Now , $V=\pi R^2{d}_R$

$\Delta V=\pi R^2\Delta d_R$

putting in above equation,

$\Delta d_A=\frac{n_1}{n_2}\frac{\Delta V}{\pi R^2}=\frac{n_1}{n_2}\pi R^{2}x$