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.`

Explanation

To solve the problem step by step, we will analyze the relationship between the apparent depth, real depth, and the volume of water drained from the cylindrical tank.

### Step 1: Define Variables
Let:
- \( d_a \) = apparent depth of water
- \( d_r \) = real depth of water
- \( R \) = radius of the cylindrical tank (diameter = \( 2R \))
- \( x \) = rate at which the apparent depth is reducing (in cm/min)
- \( n_1 \) = refractive index of air
- \( n_2 \) = refractive index of water

### Step 2: Relate Apparent Depth to Real Depth
The relationship between the apparent depth \( d_a \) and the real depth \( d_r \) can be expressed using the refractive indices:
\[ \frac{d_r}{d_a} = \frac{n_2}{n_1} \]
From this, we can express \( d_a \) in terms of \( d_r \):
\[ d_a = \frac{n_1}{n_2} d_r \]

### Step 3: Differentiate with Respect to Time
To find the rate of change of the apparent depth with respect to time, we differentiate both sides:
\[ \frac{d(d_a)}{dt} = \frac{n_1}{n_2} \frac{d(d_r)}{dt} \]
Let \( \frac{d(d_a)}{dt} = -x \) (since the apparent depth is decreasing). Then:
\[ -x = \frac{n_1}{n_2} \frac{d(d_r)}{dt} \]
This gives us:
\[ \frac{d(d_r)}{dt} = -\frac{n_2}{n_1} x \]

### Step 4: Volume of Water in the Tank
The volume \( V \) of water in the cylindrical tank is given by:
\[ V = \pi R^2 d_r \]
Differentiating this with respect to time gives:
\[ \frac{dV}{dt} = \pi R^2 \frac{d(d_r)}{dt} \]

### Step 5: Substitute for \( \frac{d(d_r)}{dt} \)
Substituting \( \frac{d(d_r)}{dt} \) from Step 3 into the volume equation:
\[ \frac{dV}{dt} = \pi R^2 \left(-\frac{n_2}{n_1} x\right) \]
Thus:
\[ \frac{dV}{dt} = -\frac{\pi R^2 n_2}{n_1} x \]

### Step 6: Amount of Water Drained
The amount of water drained in cubic centimeters per minute is given by:
\[ \frac{dV}{dt} = -\text{(amount of water drained)} \]
Thus, the amount of water drained per minute is:
\[ \text{Amount drained} = \frac{\pi R^2 n_2}{n_1} x \]

### Final Answer
The amount of water drained in \( c.c. \) per minute is:
\[ \frac{\pi R^2 n_2}{n_1} x \]