question_answer

The apparent depth of water in cylindrical water tank of diameter 2R 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) \[\frac{1}{{{v}_{0}}}=\frac{1}{200}+\frac{1}{-200\times {{10}^{3}}}\]

B) \[=\frac{{{10}^{3}}-1}{200\times {{10}^{3}}}\]    

C)       \[\frac{2\pi \,R\,{{n}_{1}}}{{{n}_{2}}}\]  

D)                       \[{{v}_{o}}=\frac{200\times {{10}^{3}}}{999}cm\]

Correct Answer: B

Solution :

                Let actual height of water of the tank be h, then \[a\propto -y\] Also  \[{{f}_{c}}=9\sqrt{{{N}_{\max }}}\] \[{{m}^{3}}\]  \[\therefore \] where x is a apparent depth. \[{{N}_{\max }}=\frac{f_{c}^{1}}{81}=\frac{{{(10\times {{10}^{6}})}^{2}}}{81}\]   \[\frac{{{n}_{2}}}{{{n}_{1}}\,}\,=\frac{\frac{dh}{dt}}{\frac{dx}{dt}}\] \[\theta ={{\sin }^{-1}}\sqrt{n_{1}^{2}-n_{2}^{2}}\]  \[e=-\frac{\Delta \phi }{\Delta t}\] or change in actual rate of flow \[=2BA\cos \phi \] change  in apparent rate of flow or         \[\theta ={{0}^{o}},\Delta t=\frac{1}{100}s\] Multiplying both sides by \[\Delta \phi =2\times 0.01\times \pi \times {{(1)}^{2}}\times 200\times \cos {{0}^{o}}\]we have \[\therefore \] \[e=\frac{-2\times 0.01\times \pi \times {{(1)}^{2}}\times 200}{100}\] Amount of water drained \[=-4\pi Volt\]