The apparent depth of ater in water in cylindrical water tank of diameter `2R cm` is reducing at the rate of `x cm//min` when is being drined out a constant rate. The amount of water drained in `cc//min` is (`n_1=`refractive index of air, `n_2=`refractive index of water)

Options

A. `(xpiR^2n_1)/n_2`
B. `(xpiR^2n_1)/n_1`
C. `(2piRn_1)/n_2`
D. `piR^2x`

Explanation

To solve the problem, we need to determine the amount of water drained from a cylindrical water tank as the apparent depth of water decreases at a certain rate. Let's break down the solution step-by-step.

### Step 1: Understand the relationship between apparent depth and actual depth
The apparent depth \( h' \) of water in a medium is related to the actual depth \( h \) by the formula:
\[ h' = \frac{h}{\mu} \]
where \( \mu \) is the refractive index of the medium. In this case, we have:
- \( n_1 \) = refractive index of air
- \( n_2 \) = refractive index of water

Thus, the relationship can be expressed as:
\[ h' = \frac{h}{n_2/n_1} = \frac{n_1 h}{n_2} \]

### Step 2: Differentiate the relationship with respect to time
To find the rate of change of apparent depth, we differentiate both sides with respect to time \( t \):
\[ \frac{dh'}{dt} = \frac{n_1}{n_2} \frac{dh}{dt} \]

### Step 3: Substitute the known rate of change
According to the problem, the apparent depth is reducing at a rate of \( x \) cm/min, which means:
\[ \frac{dh'}{dt} = -x \text{ cm/min} \]
Substituting this into the differentiated equation gives:
\[ -x = \frac{n_1}{n_2} \frac{dh}{dt} \]

### Step 4: Solve for \( \frac{dh}{dt} \)
Rearranging the equation to find \( \frac{dh}{dt} \):
\[ \frac{dh}{dt} = -\frac{n_2}{n_1} x \]

### Step 5: Calculate the volume of water drained
The volume \( V \) of water drained per minute can be calculated using the cross-sectional area of the cylindrical tank and the rate of change of height:
\[ V = A \cdot \frac{dh}{dt} \]
The cross-sectional area \( A \) of the cylinder is given by:
\[ A = \pi r^2 \]
where \( r \) is the radius of the tank. Since the diameter is \( 2R \), we have \( r = R \). Thus:
\[ A = \pi R^2 \]
Now substituting \( A \) and \( \frac{dh}{dt} \) into the volume equation:
\[ V = \pi R^2 \left(-\frac{n_2}{n_1} x\right) \]
This gives us the volume of water drained per minute:
\[ V = -\frac{n_2 \pi R^2 x}{n_1} \text{ cc/min} \]
Since volume cannot be negative, we can express it as:
\[ V = \frac{n_2 \pi R^2 x}{n_1} \text{ cc/min} \]

### Final Answer
The amount of water drained from the tank per minute is:
\[ \frac{n_2 \pi R^2 x}{n_1} \text{ cc/min} \]