A ball is dropped from a height of 20 m above the surface of water in a lake. The refractive index of water is 4.3. A fish inside the lake, in the line of fall of the ball, is looking at the ball. At an instant, when the ball is 12.8 m above the water surface, the fish sees the speed of the ball as `[Take _(g)=10m/s^(2).]`

Options

A. 9 m/s
B. 12 m/s
C. 16 m/s
D. 21.33 m/s

Explanation

To solve the problem step by step, we will follow the physics principles involved in the motion of the ball and the effect of the refractive index on the perception of speed by the fish.

### Step 1: Determine the speed of the ball just before it enters the water.

The ball is dropped from a height of 20 m, and we need to find its speed when it is at a height of 12.8 m.

Using the equation of motion:
\[ V^2 = U^2 + 2aS \]

Where:
- \( V \) = final velocity (speed of the ball at 12.8 m)
- \( U \) = initial velocity = 0 (since the ball is dropped)
- \( a \) = acceleration due to gravity = \( 10 \, \text{m/s}^2 \)
- \( S \) = displacement = initial height - current height = \( 20 \, \text{m} - 12.8 \, \text{m} = 7.2 \, \text{m} \)

Substituting the values:
\[ V^2 = 0 + 2 \times 10 \times 7.2 \]
\[ V^2 = 144 \]
\[ V = \sqrt{144} = 12 \, \text{m/s} \]

### Step 2: Calculate the perceived speed of the ball by the fish.

The speed of the ball in air is \( 12 \, \text{m/s} \). The fish is in water, and the speed of light changes when it passes from one medium to another. The perceived speed of the ball by the fish can be calculated using the refractive index.

The refractive index \( \mu \) of water is given as \( 4.3 \). The relationship between the actual speed of the ball in air and the perceived speed in water is given by:
\[ V' = \mu \cdot V \]

Where:
- \( V' \) = perceived speed of the ball by the fish
- \( V \) = actual speed of the ball = \( 12 \, \text{m/s} \)

Substituting the values:
\[ V' = 4.3 \cdot 12 \]
\[ V' = 51.6 \, \text{m/s} \]

### Final Answer:
The fish sees the speed of the ball as \( 51.6 \, \text{m/s} \).

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