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. `7.5xx10^(-3)` m
C. `9.6xx10^(-3) m`
D. `3.6xx10^(-3)` m

Explanation

To solve the problem of finding the diameter of the water stream at a distance of 0.2 m below the tap, we can follow these steps:

### Step 1: Identify Given Data
- Internal diameter of the tap, \( d_1 = 8 \times 10^{-3} \, \text{m} \)
- Velocity of water at the tap, \( V_1 = 0.4 \, \text{m/s} \)
- Height below the tap, \( h = 0.2 \, \text{m} \)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \)

### Step 2: Calculate the Velocity at Point B
To find the velocity of the water stream at point B (0.2 m below the tap), we can use the kinematic equation:

\[ V_2^2 = V_1^2 + 2gh \]

Substituting the known values:

\[ V_2^2 = (0.4)^2 + 2 \times 10 \times 0.2 \]

Calculating each term:

\[ V_2^2 = 0.16 + 4 = 4.16 \]

Now take the square root to find \( V_2 \):

\[ V_2 = \sqrt{4.16} \approx 2 \, \text{m/s} \]

### Step 3: Apply the Continuity Equation
According to the principle of conservation of mass (continuity equation), the product of the cross-sectional area and velocity at two points must be equal:

\[ A_1 V_1 = A_2 V_2 \]

Where:
- \( A_1 = \pi r_1^2 \) and \( A_2 = \pi r_2^2 \)
- \( r_1 = \frac{d_1}{2} = \frac{8 \times 10^{-3}}{2} = 4 \times 10^{-3} \, \text{m} \)

Thus, we can write:

\[ \pi (4 \times 10^{-3})^2 \times 0.4 = \pi r_2^2 \times 2 \]

The \( \pi \) cancels out:

\[ (4 \times 10^{-3})^2 \times 0.4 = r_2^2 \times 2 \]

### Step 4: Solve for \( r_2 \)
Rearranging gives:

\[ r_2^2 = \frac{(4 \times 10^{-3})^2 \times 0.4}{2} \]

Calculating the left side:

\[ r_2^2 = \frac{16 \times 10^{-6} \times 0.4}{2} = \frac{6.4 \times 10^{-6}}{2} = 3.2 \times 10^{-6} \]

Now take the square root:

\[ r_2 = \sqrt{3.2 \times 10^{-6}} \approx 1.79 \times 10^{-3} \, \text{m} \]

### Step 5: Calculate the Diameter at Point B
The diameter \( d_2 \) at point B is:

\[ d_2 = 2r_2 = 2 \times 1.79 \times 10^{-3} \approx 3.58 \times 10^{-3} \, \text{m} \]

### Final Answer
The diameter of the water stream at a distance of 0.2 m below the tap is approximately:

\[ d_2 \approx 3.58 \times 10^{-3} \, \text{m} \]