question_answer

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 cc 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) \[\pi {{R}^{2}}x\]

Correct Answer: B

Solution :

Let actual height of water of the tank be h, then \[_{1}{{n}_{2}}=\frac{actual\,\,depth}{apparent\,\,depth}\] Also \[_{1}{{n}_{2}}=\frac{{{n}_{2}}}{{{n}_{1}}}\] \[\therefore \,\,\,\,\,\frac{{{n}_{2}}}{{{n}_{1}}}=\frac{h}{x}\] where x is an apparent depth. \where x is an 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 \[\frac{dh}{dt}=\frac{{{n}_{2}}}{{{n}_{1}}}\times x\,\,cm/min\] Multiplying both sides by it \[\pi {{R}^{2}}\], we have \[\frac{dh}{dt}\times \pi {{R}^{2}}=\frac{{{n}_{2}}}{{{n}_{1}}}\times x\times \pi {{R}^{2}}\] \[\therefore \,\,\,\,\,Amount\,\,of\,\,water\,\,drained=x\,\pi \,{{R}^{2}}\frac{{{n}_{2}}}{{{n}_{1}}}\]