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. `(xpiR^(2)n_(1))/n_(2)`
B. `(xpiR^(2)n_(2))/n_(1)`
C. `(2piRn_(1))/n_(2)`
D. `piR^(2)x`

Explanation

To solve the problem, we need to find the amount of water drained from a cylindrical water tank per minute, given that the apparent depth of water is reducing at a certain rate when viewed from above. We will use the concepts of optics and geometry to derive the solution.

### Step-by-Step Solution:

1. **Understanding Apparent Depth**:
The apparent depth \( d' \) of the water in the tank is related to the actual depth \( d \) by the formula:
\[ d' = d \cdot \frac{n_1}{n_2} \]
where \( n_1 \) is the refractive index of air and \( n_2 \) is the refractive index of water.

2. **Differentiating Apparent Depth**:
Since the apparent depth is decreasing at a rate of \( x \) cm/min, we differentiate the equation:
\[ \frac{d(d')}{dt} = \frac{n_1}{n_2} \cdot \frac{d(d)}{dt} \]
This gives us:
\[ \frac{d(d')}{dt} = -x \quad \text{(since it's decreasing)} \]

3. **Relating Actual Depth Change**:
Let \( \frac{d(d)}{dt} \) be the rate of decrease of actual depth, which we denote as \( \frac{dh}{dt} \). Thus, we can write:
\[ -x = \frac{n_1}{n_2} \cdot \frac{dh}{dt} \]
Rearranging gives:
\[ \frac{dh}{dt} = -\frac{n_2}{n_1} \cdot x \]

4. **Finding the Volume Flow Rate**:
The volume flow rate \( \frac{dV}{dt} \) of water being drained can be expressed in terms of the area of the base of the cylindrical tank and the rate of change of height:
\[ \frac{dV}{dt} = A \cdot \frac{dh}{dt} \]
where \( A \) is the cross-sectional area of the tank. For a cylinder of diameter \( 2R \):
\[ A = \pi R^2 \]

5. **Substituting for \( \frac{dh}{dt} \)**:
Substituting \( \frac{dh}{dt} \) into the volume flow rate equation:
\[ \frac{dV}{dt} = \pi R^2 \cdot \left(-\frac{n_2}{n_1} \cdot x\right) \]
Since we are interested in the amount of water drained (which is positive), we take the absolute value:
\[ \frac{dV}{dt} = \frac{n_2}{n_1} \cdot x \cdot \pi R^2 \]

6. **Final Expression**:
Thus, the amount of water drained in cubic centimeters per minute is:
\[ \frac{dV}{dt} = \frac{n_2}{n_1} \cdot x \cdot \pi R^2 \]