Water is flowing through two horizontal pipes of different diameters which are connected together. The diameters of the two pipes are 3 cm and 6 cm respectively. If the speed of water in the narrower tube is `4 ms^(-1)`. Then the speed of water in the wider tube is

Options

A. `16 ms^(-1)`
B. `1 m s^(-1)`
C. `4 m s^(-1)`
D. `2 m s^(-1)`

Explanation

To solve the problem, we will use the principle of conservation of mass, which is expressed through the equation of continuity for fluids. The equation states that the product of the cross-sectional area (A) and the fluid velocity (V) at any two points in a flow must be constant.

### Step-by-Step Solution:

1. **Identify the given values:**
- Diameter of the narrower pipe (D1) = 3 cm = 0.03 m
- Diameter of the wider pipe (D2) = 6 cm = 0.06 m
- Speed of water in the narrower pipe (V1) = 4 m/s

2. **Calculate the cross-sectional areas of both pipes:**
- The area (A) of a circular pipe can be calculated using the formula:
\[ A = \pi r^2 \]
- For the narrower pipe (D1 = 0.03 m):
\[ A_1 = \pi \left(\frac{D_1}{2}\right)^2 = \pi \left(\frac{0.03}{2}\right)^2 = \pi \left(0.015\right)^2 = \pi \cdot 0.000225 \approx 0.00070686 \, \text{m}^2 \]
- For the wider pipe (D2 = 0.06 m):
\[ A_2 = \pi \left(\frac{D_2}{2}\right)^2 = \pi \left(\frac{0.06}{2}\right)^2 = \pi \left(0.03\right)^2 = \pi \cdot 0.0009 \approx 0.00282743 \, \text{m}^2 \]

3. **Apply the equation of continuity:**
- According to the equation of continuity:
\[ A_1 V_1 = A_2 V_2 \]
- Rearranging for V2 (the speed of water in the wider pipe):
\[ V_2 = \frac{A_1 V_1}{A_2} \]

4. **Substituting the values:**
- Substitute A1, V1, and A2 into the equation:
\[ V_2 = \frac{0.00070686 \cdot 4}{0.00282743} \]
- Calculate V2:
\[ V_2 \approx \frac{0.00282744}{0.00282743} \approx 1 \, \text{m/s} \]

5. **Final Answer:**
- The speed of water in the wider tube (V2) is approximately **1 m/s**.