Water is flowing continuously from a tap of area `10^(-4)m^(2)` the water velocity as it leaves the top is `1m//s` find out area of the water stream at a distance 0.15 m below the top

Options

A. `2xx10^(-5)m^(2)`
B. `5xx10^(-5)m^(2)`
C. `5xx10^(-5)m^(2)`
D. `1xx10^(-5)m^(2)`

Answer

To solve the problem of finding the area of the water stream at a distance of 0.15 m below the tap, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Values**:
- Area of the tap (A1) = \(10^{-4} \, m^2\)
- Initial velocity of water (V1) = \(1 \, m/s\)
- Height below the tap (h) = \(0.15 \, m\)
- Acceleration due to gravity (g) = \(10 \, m/s^2\) (approximate value)

2. **Use the Equation of Continuity**:
The equation of continuity states that the product of the area and velocity at one point is equal to the product of the area and velocity at another point:
\[ A_1 V_1 = A_2 V_2 \]
where \(A_2\) is the area of the stream at a distance h below the tap and \(V_2\) is the velocity at that point.

3. **Calculate the Velocity at Height h (V2)**:
To find \(V_2\), we can use the following kinematic equation:
\[ V_2 = \sqrt{V_1^2 + 2gh} \]
Substituting the known values:
\[ V_2 = \sqrt{(1)^2 + 2 \times 10 \times 0.15} \]
\[ V_2 = \sqrt{1 + 3} = \sqrt{4} = 2 \, m/s \]

4. **Substitute Values into the Continuity Equation**:
Now we can substitute \(V_1\), \(V_2\), and \(A_1\) into the continuity equation to find \(A_2\):
\[ A_2 = \frac{A_1 V_1}{V_2} \]
Substituting the values:
\[ A_2 = \frac{(10^{-4}) \times 1}{2} \]
\[ A_2 = \frac{10^{-4}}{2} = 0.5 \times 10^{-4} \, m^2 \]

5. **Convert to Standard Form**:
We can express \(A_2\) in standard form:
\[ A_2 = 5 \times 10^{-5} \, m^2 \]

### Final Answer:
The area of the water stream at a distance of 0.15 m below the tap is:
\[ A_2 = 5 \times 10^{-5} \, m^2 \]