Water flows through a horizontal pipe of radius 1 cm at a speed of 8 cm `s^(-1)`. What will be the speed of flow of water under similar conditions, if the radius of the pipe is doubled ?

Options

A. 8 cm `s^(-1)`
B. 4 cm `s^(-1)`
C. 2 cm `s^(-1)`
D. 1 cm `s^(-1)`

Answer

To solve the problem, we will use the principle of conservation of mass, which is expressed through the continuity equation for incompressible fluids. The continuity equation states that the product of the cross-sectional area of the pipe and the fluid velocity at one point is equal to the product of the cross-sectional area and the fluid velocity at another point.

### Step-by-step Solution:

1. **Identify the given values:**
- Initial radius of the pipe, \( r_1 = 1 \, \text{cm} \)
- Initial speed of water, \( v_1 = 8 \, \text{cm/s} \)
- New radius of the pipe, \( r_2 = 2 \times r_1 = 2 \, \text{cm} \)

2. **Calculate the cross-sectional areas:**
- The cross-sectional area of the pipe can be calculated using the formula:
\[ A = \pi r^2 \]
- For the initial radius:
\[ A_1 = \pi (r_1)^2 = \pi (1 \, \text{cm})^2 = \pi \, \text{cm}^2 \]
- For the new radius:
\[ A_2 = \pi (r_2)^2 = \pi (2 \, \text{cm})^2 = 4\pi \, \text{cm}^2 \]

3. **Apply the continuity equation:**
- According to the continuity equation:
\[ A_1 v_1 = A_2 v_2 \]
- Substituting the areas and known values:
\[ \pi v_1 = 4\pi v_2 \]

4. **Simplify the equation:**
- Cancel out \( \pi \) from both sides:
\[ v_1 = 4 v_2 \]

5. **Solve for \( v_2 \):**
- Rearranging the equation gives:
\[ v_2 = \frac{v_1}{4} \]
- Substituting the value of \( v_1 \):
\[ v_2 = \frac{8 \, \text{cm/s}}{4} = 2 \, \text{cm/s} \]

### Final Answer:
The speed of flow of water when the radius of the pipe is doubled is \( v_2 = 2 \, \text{cm/s} \).