Water is flowing continuously from a tap having an internal diameter `8xx10^(-3)`m. The water velocity as it leves the tap is `0.4 ms^(-1)`. The diameter of the water stream at a distance `2xx10^(-1)`m below the tap is close to `(g=10m//s^(2))`

Options

A. ` 5.0 xx 10 ^(-3) m `
B. ` 7.5 xx 10 ^(-3 ) m `
C. `9.6 xx 10 ^(-3) m `
D. `3.6 xx 10 ^(-3 ) m `

Explanation

To solve the problem step by step, we will follow the principles of fluid dynamics and the equations of motion.

### Step 1: Identify the given data
- Internal diameter of the tap, \( d_1 = 8 \times 10^{-3} \) m
- Velocity of water leaving the tap, \( v_1 = 0.4 \, \text{m/s} \)
- Height below the tap, \( h = 2 \times 10^{-1} \) m
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \)

### Step 2: Calculate the velocity of the water stream at a distance \( h \) below the tap
We can use the equation of motion for a freely falling object to find the velocity \( v_2 \) at a height \( h \) below the tap:

\[ v_2 = \sqrt{v_1^2 + 2gh} \]

Substituting the values:

\[ v_2 = \sqrt{(0.4)^2 + 2 \times 10 \times (2 \times 10^{-1})} \]

Calculating this step-by-step:

1. Calculate \( (0.4)^2 = 0.16 \)
2. Calculate \( 2 \times 10 \times (2 \times 10^{-1}) = 2 \times 10 \times 0.2 = 4 \)
3. Now, sum these values: \( 0.16 + 4 = 4.16 \)
4. Finally, take the square root:

\[ v_2 = \sqrt{4.16} \approx 2.04 \, \text{m/s} \quad (\text{approximately } 2 \, \text{m/s}) \]

### Step 3: Use the principle of continuity to find the diameter of the stream
According to the principle of continuity, the product of the cross-sectional area and the velocity at two points must be equal:

\[ A_1 v_1 = A_2 v_2 \]

Where:
- \( A_1 = \frac{\pi d_1^2}{4} \)
- \( A_2 = \frac{\pi d_2^2}{4} \)

Substituting these into the equation:

\[ \frac{\pi (8 \times 10^{-3})^2}{4} \cdot 0.4 = \frac{\pi d_2^2}{4} \cdot 2 \]

### Step 4: Simplify the equation
Cancelling \( \frac{\pi}{4} \) from both sides:

\[ (8 \times 10^{-3})^2 \cdot 0.4 = d_2^2 \cdot 2 \]

Calculating \( (8 \times 10^{-3})^2 \):

\[ (8 \times 10^{-3})^2 = 64 \times 10^{-6} = 6.4 \times 10^{-5} \]

Now substituting back:

\[ 6.4 \times 10^{-5} \cdot 0.4 = d_2^2 \cdot 2 \]

Calculating \( 6.4 \times 10^{-5} \cdot 0.4 \):

\[ 6.4 \times 0.4 = 2.56 \times 10^{-5} \]

So we have:

\[ 2.56 \times 10^{-5} = d_2^2 \cdot 2 \]

### Step 5: Solve for \( d_2^2 \)
Dividing both sides by 2:

\[ d_2^2 = \frac{2.56 \times 10^{-5}}{2} = 1.28 \times 10^{-5} \]

### Step 6: Calculate \( d_2 \)
Taking the square root:

\[ d_2 = \sqrt{1.28 \times 10^{-5}} \approx 3.6 \times 10^{-3} \, \text{m} \]

### Conclusion
The diameter of the water stream at a distance \( 2 \times 10^{-1} \) m below the tap is approximately \( 3.6 \times 10^{-3} \, \text{m} \).

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