Water is flowing continuously from a tap having an internal diameter `8xx10^-3m`. The water velocity as it leaves the tap is `0.4ms^-1`. The diameter of the water stream at a distance `2xx10^-1m` below the tap is close to:

Options

A. (a) `7.5xx10^-3m`
B. (b) `9.6xx10^-3m`
C. (c) `3.6xx10^-3m`
D. (d) `5.0xx10^-3m`

Explanation

To find the diameter of the water stream at a distance of \(2 \times 10^{-1} \, m\) below the tap, we can use the principle of conservation of mass, specifically the equation of continuity for incompressible fluids. The equation states that the product of the cross-sectional area and the velocity of the fluid must remain constant along the flow.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Internal diameter of the tap, \(d_1 = 8 \times 10^{-3} \, m\)
- Velocity of water at the tap, \(v_1 = 0.4 \, m/s\)
- Distance below the tap, \(h = 2 \times 10^{-1} \, m\)

2. **Calculate the Cross-Sectional Area at the Tap:**
The cross-sectional area \(A_1\) of the tap can be calculated using the formula for the area of a circle:
\[ A_1 = \pi \left(\frac{d_1}{2}\right)^2 = \pi \left(\frac{8 \times 10^{-3}}{2}\right)^2 \]
\[ A_1 = \pi \left(4 \times 10^{-3}\right)^2 = \pi \times 16 \times 10^{-6} \, m^2 = 16\pi \times 10^{-6} \, m^2 \]

3. **Apply the Equation of Continuity:**
According to the equation of continuity:
\[ A_1 v_1 = A_2 v_2 \]
where \(A_2\) is the cross-sectional area of the stream at a distance \(h\) below the tap, and \(v_2\) is the velocity at that point.

4. **Assuming the Velocity at \(h\) is the Same:**
For simplicity, we can assume that the velocity of the water stream remains approximately the same at \(h\) (since the height is not very large compared to the diameter). Thus, \(v_2 \approx v_1 = 0.4 \, m/s\).

5. **Calculate the Cross-Sectional Area at \(h\):**
Rearranging the equation of continuity gives:
\[ A_2 = \frac{A_1 v_1}{v_2} = A_1 \]
Therefore, \(A_2\) remains the same as \(A_1\) under our assumption.

6. **Calculate the Diameter at \(h\):**
Since \(A_2\) is the same, we can find the diameter \(d_2\) using:
\[ A_2 = \pi \left(\frac{d_2}{2}\right)^2 \]
Setting \(A_2 = A_1\):
\[ 16\pi \times 10^{-6} = \pi \left(\frac{d_2}{2}\right)^2 \]
Dividing both sides by \(\pi\):
\[ 16 \times 10^{-6} = \left(\frac{d_2}{2}\right)^2 \]
Taking the square root:
\[ \frac{d_2}{2} = 4 \times 10^{-3} \implies d_2 = 8 \times 10^{-3} \, m \]

7. **Conclusion:**
The diameter of the water stream at a distance \(2 \times 10^{-1} \, m\) below the tap is approximately \(8 \times 10^{-3} \, m\).