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 `c.c.` per minute is `(n_(1)=` refractive index of air, `n_(2)=` refractive index of water )

Options

A. `(x pi R^(2) n_(1))/(n_(2))`
B. `(x pi R^(2)n_(2))/(n_(1))`
C. `(2 pi R n_(1))/(n_(2))`
D. `pi R^(2)x.`

Answer

B

Explanation

`(n_(2))/(n_(1))=("real depth")/("apparent depth")" or, "(n_(2))/(n_(1))=(h)/(x)`
Differentiating with respect to time, ltbtgt `(n_(2))/(n_(1))=(dh//dt)/(dx//dt)`
Change in real depth `=(n_(2))/(n_(1))xx` change in apparent depth `(dh)/(dt)=(n_(2))/(n_(1))x` cm/min.
The amount of water drained in c.c. per minute ltbtgt `=(dh)/(dt)xxpi R^(2)=x pi R^(2)(n_(2))/(n_(1))`.