Water from a tap emerges vertically downward with an initial speed of 3.0 m/s. The cross-sectional area of the tap is `10^(-4)m^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 2 m below the tap is:

Options

A. `0.7 xx 10^(-5) m^(2)`
B. `10^(-4) m^(2)`
C. `4.3 xx 10^(-5) m^(2)`
D. `2.8 xx 10^(-5) m^(2)`

Explanation

To solve the problem, we will use the principles of fluid dynamics, specifically the continuity equation and Bernoulli's equation.

### Step-by-Step Solution:

1. **Identify Given Data:**
- Initial speed of water, \( v_1 = 3.0 \, \text{m/s} \)
- Cross-sectional area of the tap, \( A_1 = 10^{-4} \, \text{m}^2 \)
- Height below the tap, \( h = 2 \, \text{m} \)

2. **Use Bernoulli's Equation:**
Bernoulli's equation states that for an incompressible fluid flowing in a streamline, the following holds:
\[ P_1 + \frac{1}{2} \rho v_1^2 + \rho g h_1 = P_2 + \frac{1}{2} \rho v_2^2 + \rho g h_2 \]
Since the pressure is constant throughout the stream, we can simplify this equation. We can set \( h_1 = 0 \) (at the tap) and \( h_2 = -2 \, \text{m} \) (2 m below the tap).

Thus, the equation simplifies to:
\[ P + \frac{1}{2} \rho (3.0)^2 = P + \frac{1}{2} \rho v_2^2 - 2 \rho g \]

3. **Cancel out the pressure terms:**
Since \( P_1 = P_2 \), we can eliminate \( P \) from both sides:
\[ \frac{1}{2} \rho (3.0)^2 = \frac{1}{2} \rho v_2^2 - 2 \rho g \]

4. **Rearranging the equation:**
\[ \frac{1}{2} \rho v_2^2 = \frac{1}{2} \rho (3.0)^2 + 2 \rho g \]
Dividing through by \( \rho \) (assuming it is non-zero):
\[ \frac{1}{2} v_2^2 = \frac{1}{2} (3.0)^2 + 2g \]

5. **Substituting \( g = 9.81 \, \text{m/s}^2 \):**
\[ \frac{1}{2} v_2^2 = \frac{1}{2} (9) + 2(9.81) \]
\[ \frac{1}{2} v_2^2 = 4.5 + 19.62 = 24.12 \]

6. **Solving for \( v_2 \):**
\[ v_2^2 = 48.24 \implies v_2 = \sqrt{48.24} \approx 6.96 \, \text{m/s} \]

7. **Using the Continuity Equation:**
The continuity equation states that:
\[ A_1 v_1 = A_2 v_2 \]
Rearranging gives:
\[ A_2 = \frac{A_1 v_1}{v_2} \]

8. **Substituting the known values:**
\[ A_2 = \frac{10^{-4} \times 3.0}{6.96} \approx \frac{3.0 \times 10^{-4}}{6.96} \approx 4.31 \times 10^{-5} \, \text{m}^2 \]

### Final Answer:
The cross-sectional area of the stream 2 m below the tap is approximately \( 4.31 \times 10^{-5} \, \text{m}^2 \).