Question

The apparent depth of water in a cylindrical water tank of diameter $2R\text{cm}$ is reducing at the rate of $x\text{cm/min}$ when water is being drained out at a constant rate. The actual amount of water drained in ${\text{cm}}^3\text{/min}$ 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

The correct option is B $\frac{x\pi R^2{n}_2}{n_1}$

On applying,
$\frac{{\mu }_2}{{\mu }_1}=\frac{d_{real}}{d_{app}}$
$\Rightarrow \frac{n_2}{n_1}=\frac{d_{real}}{d_{app}}$
or, $d_{app}=\left(\frac{n_1}{n_2}\right)d_{real}$
On differentiating both sides $w.r.t$ time,
$\frac{d\left(d_{app}\right)}{dt}=\left(\frac{n_1}{n_2}\right)\frac{d\left(d_{real}\right)}{dt}$
The rate of reduction of apparent depth is,
$\frac{d\left(d_{app}\right)}{dt}=-x\text{cm/min}$
$\Rightarrow \frac{d\left(d_{real}\right)}{dt}=-\frac{n_{2}x}{n_1}$ ...............$\left(i\right)$
Rate of drainage of water will be,
$\frac{-dV}{dt}=\frac{-d}{dt}\left(A{d}_{real}\right)$
$\Rightarrow \frac{-dV}{dt}=-A\frac{d\left(d_{real}\right)}{dt}$
$\Rightarrow \frac{-dV}{dt}=-\pi R^2\left(\frac{-n_{2}x}{n_1}\right)$
$[$from $\left(i\right)]$
$\therefore \frac{-dV}{dt}=\frac{x\pi R^2{n}_2}{n_1}\text{cc/min}$

Why this question? Tips: Apply the concept of real depth and apparent depth and use differentiation to get rate of volumetric change.