Water from a tap of cross sectional area 1`cm^2` starts falling down vertically, with a speed of `1m//s`. What is the area of cross section of the stream of water at a distance of 20cm below the mouth of the tap?
[Assume that (1) the flow is steady, (2) pressure is constant throughout the stream of water and g=`10m//s^2`]

Options

A. `3xx10^(-5)m^2`
B. `4xx10^(-5)m^2`
C. `5xx10^(-5)m^2`
D. `6xx10^(-5)m^2`

Explanation

To solve the problem, we will use the principle of conservation of mass, which is represented by the continuity equation. The continuity equation states that the product of the cross-sectional area and the velocity of fluid flow must remain constant along the streamline.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Initial cross-sectional area of the tap, \( A_1 = 1 \, \text{cm}^2 = 1 \times 10^{-4} \, \text{m}^2 \)
- Initial velocity of water from the tap, \( v_1 = 1 \, \text{m/s} \)
- Distance below the tap, \( h = 20 \, \text{cm} = 0.2 \, \text{m} \)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \)

2. **Calculate the Velocity of Water at 20 cm Below the Tap:**
- Using the equation of motion under gravity:
\[ v_2 = v_1 + \sqrt{2gh} \]
- Substitute the values:
\[ v_2 = 1 + \sqrt{2 \times 10 \times 0.2} \]
\[ v_2 = 1 + \sqrt{4} = 1 + 2 = 3 \, \text{m/s} \]

3. **Apply the Continuity Equation:**
- According to the continuity equation:
\[ A_1 v_1 = A_2 v_2 \]
- Rearranging for \( A_2 \):
\[ A_2 = \frac{A_1 v_1}{v_2} \]

4. **Substitute the Known Values:**
- Substitute \( A_1 \), \( v_1 \), and \( v_2 \):
\[ A_2 = \frac{(1 \times 10^{-4}) \times 1}{3} \]
\[ A_2 = \frac{1 \times 10^{-4}}{3} \approx 3.33 \times 10^{-5} \, \text{m}^2 \]

5. **Convert \( A_2 \) to cm²:**
- To convert \( A_2 \) back to cm²:
\[ A_2 = 3.33 \times 10^{-5} \, \text{m}^2 \times 10^4 \, \text{cm}^2/\text{m}^2 \approx 0.333 \, \text{cm}^2 \]

### Final Answer:
The area of cross-section of the stream of water at a distance of 20 cm below the mouth of the tap is approximately **0.333 cm²**.