An observer observer a fish moving upwards in a cylindrical container of cross section area `1 m^(2)` filled with water up to a height of `5 m`. A hole is present at the botton of the container having cross section areea `1//1000 m^(2)`. Find out the speed of the image of fish observed by observer when the botton hole is just opened. (Given: The fish is moving with the speed of `6 m//s` towards the observer, `mu` of water `=4//3`)

Explanation

To solve the problem, we need to determine the speed of the image of the fish as observed by the observer when the hole at the bottom of the cylindrical container is opened. Here's a step-by-step breakdown of the solution:

### Step 1: Understanding the Problem
We have a cylindrical container filled with water, and a fish is moving upwards at a speed of 6 m/s. When the hole at the bottom is opened, water will flow out, and we need to find the speed of the image of the fish as seen by the observer.

### Step 2: Determine the Speed of Water Flowing Out
Using Bernoulli's theorem, we can find the speed of water flowing out of the hole at the bottom of the container. The speed \( v_1 \) of the water flowing out can be calculated using the formula:
\[ v_1 = \sqrt{2gh} \]
where \( g \) is the acceleration due to gravity (approximately \( 10 \, \text{m/s}^2 \)) and \( h \) is the height of the water column (5 m).

Calculating \( v_1 \):
\[ v_1 = \sqrt{2 \times 10 \times 5} = \sqrt{100} = 10 \, \text{m/s} \]

### Step 3: Apply the Continuity Equation
According to the continuity equation, the product of the area and velocity must be constant. Therefore, we can use:
\[ A_1 v_1 = A_2 v_2 \]
where:
- \( A_1 = \frac{1}{1000} \, \text{m}^2 \) (area of the hole)
- \( A_2 = 1 \, \text{m}^2 \) (cross-sectional area of the container)
- \( v_2 \) is the speed of the water flowing out of the hole.

Substituting the values:
\[ \frac{1}{1000} \times 10 = 1 \times v_2 \]
\[ v_2 = \frac{10}{1000} = 0.01 \, \text{m/s} \]

### Step 4: Calculate the Speed of the Fish with Respect to the Surface
The speed of the fish with respect to the surface of the water can be calculated as:
\[ v_{f/s} = v_f + v_2 \]
where \( v_f = 6 \, \text{m/s} \) (speed of the fish) and \( v_2 = 0.01 \, \text{m/s} \).

Thus,
\[ v_{f/s} = 6 + 0.01 = 6.01 \, \text{m/s} \]

### Step 5: Adjust for the Refractive Index
The speed of the image of the fish as observed by the observer must be adjusted for the refractive index \( \mu \) of water, which is given as \( \frac{4}{3} \). The speed of the image \( v_{image} \) can be calculated using:
\[ v_{image} = \frac{v_{f/s}}{\mu} \]
Substituting the values:
\[ v_{image} = \frac{6.01}{\frac{4}{3}} = 6.01 \times \frac{3}{4} = 4.5075 \, \text{m/s} \]

### Step 6: Final Calculation for the Speed of the Image
The final speed of the image of the fish as observed by the observer is:
\[ v_{image} = 4.5075 \, \text{m/s} \]

### Conclusion
The speed of the image of the fish observed by the observer when the bottom hole is just opened is approximately \( 4.5075 \, \text{m/s} \).

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