Water is flowing in a pipe of diameter 4 cm with a velocity `3 m//s` . The water then enters into a tube of diameter 2 cm . The velocity of water in the other pipe is

Options

A. `3 m//s`
B. `6 m//s`
C. `12 m//s`
D. `8 m//s`

Answer

To solve the problem, we will use the principle of continuity for fluid flow, which states that the product of the cross-sectional area and the velocity of flow must remain constant along a streamline.

### Step-by-Step Solution:

1. **Identify the given values:**
- Diameter of the first pipe, \( D_1 = 4 \, \text{cm} \)
- Velocity in the first pipe, \( V_1 = 3 \, \text{m/s} \)
- Diameter of the second pipe, \( D_2 = 2 \, \text{cm} \)

2. **Convert diameters to meters:**
- \( D_1 = 4 \, \text{cm} = 0.04 \, \text{m} \)
- \( D_2 = 2 \, \text{cm} = 0.02 \, \text{m} \)

3. **Calculate the cross-sectional areas of both pipes:**
- The area \( A \) of a circle is given by the formula:
\[ A = \frac{\pi D^2}{4} \]
- For the first pipe:
\[ A_1 = \frac{\pi (0.04)^2}{4} = \frac{\pi (0.0016)}{4} = 0.00125664 \, \text{m}^2 \]
- For the second pipe:
\[ A_2 = \frac{\pi (0.02)^2}{4} = \frac{\pi (0.0004)}{4} = 0.00031416 \, \text{m}^2 \]

4. **Apply the continuity equation:**
- According to the continuity equation:
\[ A_1 V_1 = A_2 V_2 \]
- Rearranging for \( V_2 \):
\[ V_2 = \frac{A_1 V_1}{A_2} \]

5. **Substituting the values:**
- Substitute \( A_1 \), \( V_1 \), and \( A_2 \):
\[ V_2 = \frac{(0.00125664) \times (3)}{0.00031416} \]
- Calculate \( V_2 \):
\[ V_2 = \frac{0.00376992}{0.00031416} \approx 12 \, \text{m/s} \]

6. **Final Answer:**
- The velocity of water in the second pipe is \( V_2 = 12 \, \text{m/s} \).