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 $
