Optics Question 135

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 (n1 = refractive index of air, n2 = refractive index of water)

[AIIMS 2005]

Options:

A) x p R2 n1/n2

B) x p R2 n2/n1

C) 2 p R n1/n2

D) p R2 x

Show Answer

Answer:

Correct Answer: B

Solution:

Apparent depth $ h’=\frac{h}{ _{air}{\mu _{liquid}}} $

Therefore $ \frac{dh’}{dt}=\frac{1}{ _{a}{\mu _{w}}}=\frac{1}{ _{a}{\mu _{w}}}\frac{dh}{dt} $

Therefore $ x=\frac{1}{ _{a}{\mu _{w}}}\frac{dh}{dt} $

Therefore $ \frac{dh}{dt}= _{a}{\mu _{w}}x $ Now volume of water $ V=\pi R^{2}h $

Therefore $ \frac{dV}{dt} $

$ =\pi R^{2}\frac{dh}{dt} $

$ =\pi R^{2}. _{a}{\mu _{w}}x $

$ = _{a}{\mu _{w}}\pi R^{2}x $

$ =\frac{{\mu _{w}}}{{\mu _{a}}}\pi R^{2}x $

$ =( \frac{n _{2}}{n _{1}} )\pi R^{2}x $



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