Water from a tap emerges vertically downwards with an initial spped of `1.0ms^-1`. The cross-sectional area of the tap is `10^-4m^2`. Assume that the pressure is constant throughout the stream of water, and that the flow is steady. The cross-sectional area of the stream 0.15 m below the tap is

Options

A. `5.0 xx 10^(-4)m^(2)`
B. `1.0 xx 10^(-4) m^(2)`
C. `5.0 xx 10^(-5)m^(2)`
D. `2.0 xx 10^(-5)m^(2)`

Explanation

To solve the problem, we will use the principles of fluid mechanics, specifically the conservation of energy and the continuity equation.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Initial speed of water, \( v_1 = 1.0 \, \text{m/s} \)
- Cross-sectional area of the tap, \( A_1 = 10^{-4} \, \text{m}^2 \)
- Height below the tap, \( h = 0.15 \, \text{m} \)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \) (assuming standard value)

2. **Apply Conservation of Energy:**
The total mechanical energy at the tap (point A) must equal the total mechanical energy at a point 0.15 m below (point B). The potential energy at point A is converted into kinetic energy at point B.

The energy equation can be written as:
\[ \text{PE}_A + \text{KE}_A = \text{PE}_B + \text{KE}_B \]
Where:
- \(\text{PE} = \rho g h\) (potential energy)
- \(\text{KE} = \frac{1}{2} \rho v^2\) (kinetic energy)

At point A:
\[ \text{PE}_A = \rho g h + \frac{1}{2} \rho v_1^2 \]
At point B (where potential energy is zero):
\[ \text{PE}_B = 0 + \frac{1}{2} \rho v_2^2 \]

Setting them equal:
\[ \rho g h + \frac{1}{2} \rho v_1^2 = \frac{1}{2} \rho v_2^2 \]

3. **Simplify the Equation:**
We can cancel \(\rho\) from both sides (assuming it is constant):
\[ g h + \frac{1}{2} v_1^2 = \frac{1}{2} v_2^2 \]

4. **Substitute Known Values:**
Substitute \(g = 10 \, \text{m/s}^2\), \(h = 0.15 \, \text{m}\), and \(v_1 = 1.0 \, \text{m/s}\):
\[ 10 \cdot 0.15 + \frac{1}{2} (1.0)^2 = \frac{1}{2} v_2^2 \]
\[ 1.5 + 0.5 = \frac{1}{2} v_2^2 \]
\[ 2 = \frac{1}{2} v_2^2 \]

5. **Solve for \(v_2\):**
Multiply both sides by 2:
\[ 4 = v_2^2 \]
Taking the square root:
\[ v_2 = 2 \, \text{m/s} \]

6. **Apply the Continuity Equation:**
According to the continuity equation:
\[ A_1 v_1 = A_2 v_2 \]
Rearranging gives:
\[ A_2 = \frac{A_1 v_1}{v_2} \]

7. **Substitute Known Values:**
\[ A_2 = \frac{10^{-4} \cdot 1.0}{2} = \frac{10^{-4}}{2} = 5 \times 10^{-5} \, \text{m}^2 \]

### Final Answer:
The cross-sectional area of the stream 0.15 m below the tap is \( A_2 = 5 \times 10^{-5} \, \text{m}^2 \).

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