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. (a) `5.5xx10^-4m^2`
B. (b) `1.0xx10^-5m^2`
C. (c) `5.0xx10^-5m^2
D. (d) `2.0xx10^-5m^2

Explanation

To solve the problem, we will use the principles of fluid dynamics, specifically the continuity equation and the equations of motion under gravity.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Initial speed of water from the tap, \( V_1 = 1.0 \, \text{m/s} \)
- 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, \( g = 9.81 \, \text{m/s}^2 \) (approximately \( 10 \, \text{m/s}^2 \) for calculations)

2. **Calculate the Speed of Water at 0.15 m Below the Tap:**
We can use the equation of motion to find the speed of the water at a height of 0.15 m below the tap. The formula is:
\[ V_2 = V_1 + \sqrt{2gh} \]
Here, \( V_1 \) is the initial speed, and \( h \) is the height.

Plugging in the values:
\[ V_2 = 1.0 + \sqrt{2 \times 10 \times 0.15} \]
\[ V_2 = 1.0 + \sqrt{3} \approx 1.0 + 1.732 \approx 2.732 \, \text{m/s} \]

3. **Use the Continuity Equation:**
The continuity equation states that the product of the cross-sectional area and the velocity at one point is equal to the product at another point:
\[ A_1 V_1 = A_2 V_2 \]
Rearranging for \( A_2 \):
\[ A_2 = \frac{A_1 V_1}{V_2} \]

4. **Substituting the Known Values:**
\[ A_2 = \frac{(10^{-4}) \times (1.0)}{2.732} \]
\[ A_2 \approx \frac{10^{-4}}{2.732} \approx 3.66 \times 10^{-5} \, \text{m}^2 \]

5. **Final Result:**
The cross-sectional area of the stream 0.15 m below the tap is approximately \( 3.66 \times 10^{-5} \, \text{m}^2 \).