Question

The apparent depth of water in cylindrical water tanks of diameter 2R cm is reducing at the rate of \[xcm{\min ^{ - 1}}\] 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)
A. \[\dfrac{{x\pi {R^2}{n_1}}}{{{n_2}}}\]
B. \[\dfrac{{x\pi {R^2}{n_2}}}{{{n_1}}}\]
C. \[\dfrac{{2\pi R{n_1}}}{{{n_2}}}\]
D. \[x\pi {R^2}\]

Answer 1

Hint We know that the ratio of the actual depth to the apparent depth of a water vessel is given as the ratio of the refractive index of water to that of air. Use this relation to find actual depth and then differentiate the equation to find rate of change of actual depth.

Complete step by step solution
Radius of the cylindrical tank is given as R
Let the height of the tank be h
Then \[Volume\,of\,water\,in\,the\,cylindrical\,\tan k = \pi {R^2}h\]
As we know that \[\dfrac{{Actual\,depth}}{{Apparent\,depth}} = \dfrac{{{n_2}}}{{{n_1}}}\]
⇒ \[Actual\,depth = \dfrac{{{n_2}}}{{{n_1}}} \times Apparent\,depth\]
Therefore, \[Rate\,of\,change\,in\,Actual\,depth = \dfrac{{{n_2}}}{{{n_1}}} \times Rate\,of\,change\,in\,Apparent\,depth\]
And the rate of change in apparent depth is given as \[xcm{\min ^{ - 1}}\]
So \[\;Rate\,of\,change\,inactual\,depth = \dfrac{{{n_2}}}{{{n_1}}} \times x\]
The amount of water drained per min =Rate of change in the Volume of water
 \[\dfrac{{dV}}{{dt}} = \dfrac{{d\left( {A \times h} \right)}}{{dt}} = \dfrac{{d\left( {\pi {R^2}h} \right)}}{{dt}}\]
Area of the cylinder is constant, therefore \[\dfrac{{dV}}{{dt}} = \pi {R^2}\dfrac{{dh}}{{dt}}\]
So, \[\dfrac{{dV}}{{dt}} = \pi {R^2} \times (Rate\,of\,change\,in\,actual\,depth) = \pi {R^2} \times \left( {\dfrac{{\;{n_2}}}{{{n_1}}} \times x} \right)\]

Hence the answer is, \[\dfrac{{dV}}{{dt}} = \dfrac{{x\pi {R^2}{n_2}}}{{{n_1}}}\]

Note We use \[\dfrac{{Actual\,depth}}{{Apparent\,depth}} = \dfrac{{{n_2}}}{{{n_1}}}\] formula to relate apparent depth and real depth as we are given the rate of change of apparent depth. And then we use the relation between the rate of change of Volume and rate of change of real height of the cylinder that is \[\dfrac{{dV}}{{dt}} = \dfrac{{d\left( {A \times h} \right)}}{{dt}} = \dfrac{{d\left( {\pi {R^2}h} \right)}}{{dt}}\] and hence derived the formula for amount of water draining per min as \[\dfrac{{dV}}{{dt}} = \dfrac{{x\pi {R^2}{n_2}}}{{{n_1}}}\]