The apparent depth of water in cylindrical water tank of diameter 2-R cm is reducing at the rate of x cm/min 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)

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${xπR}^{2} n_{1} / n_{2}$

${xπR}^{2} n_{2} / n_{1}$

$2 {πRn}_{1} / n_{2}$

${πR}^{2} x$

answer is B.

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Detailed Solution

$\begin{array}{l} Apparent depth h' = \frac{h}{_{air} μ_{liquid}} \Rightarrow \frac{dh'}{dt} = \frac{1}{_{a} μ_{w}} \frac{dh}{dt} \\ \Rightarrow x = \frac{1}{_{a} μ_{w}} \frac{dh}{dt} \Rightarrow \frac{dh}{dt} =_{a} μ_{w} x \\ Now volume of water V = {πR}^{2} h \Rightarrow \frac{dV}{dt} = {πR}^{2} \frac{dh}{dt} = {πR}^{2} ._{a} μ_{w} x \\ =_{a} μ_{w} {πR}^{2} x = \frac{μ_{w}}{μ_{o}} {πR}^{2} x = (\frac{n_{2}}{n_{1}}) {πR}^{2} x \end{array}$