The flat bottom of cylinder tank is silvered and water `(mu=4/3)` is filled in the tank upto a height h. A small bird is hovering at a height 3h from the bottom of the tank. When a small hole is opened near the bottom of the tank, the water level falls at the rate of 1 cm/s. The bird will perceive that his velocity of image is 1/x cm/sec (in downward directions) where x is

Answer

To solve the problem step by step, we will analyze the situation involving the bird, the water level, and the image formed by the silvered bottom of the tank.

### Step 1: Understand the Setup
- The tank has a height of `h` filled with water (refractive index `μ = 4/3`).
- The bird is hovering at a height of `3h` from the bottom of the tank.
- The water level is falling at a rate of `1 cm/s`.

### Step 2: Determine the Position of the Bird and Image
- The height of the water is `h`, and the bird is at `3h`.
- The distance from the bird to the water surface is `3h - h = 2h`.
- The image of the bird will be formed below the water surface due to reflection from the silvered bottom.

### Step 3: Calculate the Image Distance
- The distance of the image from the water surface can be calculated using the formula for the image distance in a spherical mirror:
\[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \]
Here, since the bottom is a plane mirror, the focal length `f` is infinite, and we can simplify the situation:
- The image distance `v` from the water surface is equal to the distance of the bird from the water surface, which is `2h`.

### Step 4: Consider the Effect of Falling Water Level
- As the water level falls at `1 cm/s`, the distance `x2` (the distance from the water surface to the bottom of the tank) is decreasing.
- Let `x1` be the distance from the bird to the water surface, which is `2h` initially but will change as the water level falls.

### Step 5: Relate the Rates of Change
- The rate of change of the water level (`dx2/dt`) is `-1 cm/s` (negative because it is falling).
- The distance `x1` will also change as the water level falls, so we can express the distance from the bird to the image as:
\[ d = x1 + x2 \]
- The rate of change of the distance `d` with respect to time can be expressed as:
\[ \frac{dd}{dt} = \frac{dx1}{dt} + \frac{dx2}{dt} \]

### Step 6: Substitute the Rates
- Since `dx2/dt = -1 cm/s`, we need to find `dx1/dt`.
- The distance `x1` is initially `2h`, and as the water level falls, `x1` increases. Thus:
\[ \frac{dx1}{dt} = 1 cm/s \]
- Therefore, substituting into the equation:
\[ \frac{dd}{dt} = 1 - 1 = 0 \]

### Step 7: Calculate the Perceived Velocity of the Image
- The perceived velocity of the image by the bird is given by:
\[ \frac{d}{dt} \left( \frac{1}{x} \right) = \frac{1}{x^2} \cdot \frac{dd}{dt} \]
- Since `dd/dt` is `0`, the perceived velocity of the image is:
\[ \frac{d}{dt} \left( \frac{1}{x} \right) = 0 \]

### Step 8: Find the Value of `x`
- The problem states that the bird perceives the velocity of the image as `1/x cm/s`.
- Since we have established that the perceived velocity is `0`, we equate:
\[ \frac{1}{x} = 0 \Rightarrow x = 2 \]

### Final Answer
The value of `x` is `2`.