What flows through a horizontal pipe of radius `r` at a speed V. if the radius of the pipe is doubled, the speed of flow of water under similar conditions is

Options

A. 2V
B. `(V)/(2)`
C. `(V)/(4)`
D. `4V`

Answer

To solve the problem, we can 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 (A) of the pipe and the velocity (V) of the fluid at any point along the pipe remains constant.

### Step-by-Step Solution:

1. **Identify the initial conditions:**
- Let the initial radius of the pipe be \( r \).
- The initial velocity of the fluid is \( V \).
- The cross-sectional area \( A_1 \) of the pipe can be calculated using the formula for the area of a circle:
\[ A_1 = \pi r^2 \]

2. **Apply the continuity equation:**
- According to the continuity equation:
\[ A_1 V_1 = A_2 V_2 \]
- Here, \( V_1 \) is the initial velocity \( V \) and \( A_2 \) is the area when the radius is doubled.

3. **Calculate the new conditions:**
- If the radius of the pipe is doubled, the new radius \( R \) becomes \( 2r \).
- The new cross-sectional area \( A_2 \) is:
\[ A_2 = \pi (2r)^2 = \pi (4r^2) = 4\pi r^2 \]

4. **Set up the equation using the continuity equation:**
- Substitute the values into the continuity equation:
\[ A_1 V_1 = A_2 V_2 \]
\[ (\pi r^2)(V) = (4\pi r^2)(V_2) \]

5. **Simplify the equation:**
- Cancel \( \pi r^2 \) from both sides:
\[ V = 4 V_2 \]

6. **Solve for the new velocity \( V_2 \):**
- Rearranging gives:
\[ V_2 = \frac{V}{4} \]

### Final Answer:
The speed of flow of water when the radius of the pipe is doubled is \( \frac{V}{4} \).