Water is flowing in a horizontal pipe ofnonuniform cross section. The velocities of water at points A and Bin the tube, are in the ratio of 4 : 1, what is the ratio of the diameters of the pipe at A and B ?

Options

A. `1:1`
B. `1:2:
C. `2:1`
D. `2:3`

Explanation

To solve the problem of finding the ratio of the diameters of the pipe at points A and B, we can use the principle of conservation of mass, which is described by the equation of continuity for fluid flow. The equation states that the product of the cross-sectional area (A) and the velocity (v) of the fluid at any two points in a flow system must be constant.

### Step-by-Step Solution:

1. **Understand the Given Information:**
- Let the velocity of water at point A be \( v_A \) and at point B be \( v_B \).
- According to the problem, the ratio of the velocities is given as:
\[ \frac{v_A}{v_B} = \frac{4}{1} \]
- This implies:
\[ v_A = 4v_B \]

2. **Apply the Equation of Continuity:**
- The equation of continuity states:
\[ A_A v_A = A_B v_B \]
- Where \( A_A \) and \( A_B \) are the cross-sectional areas at points A and B respectively.

3. **Express the Areas in Terms of Diameter:**
- The area \( A \) of a circular cross-section can be expressed in terms of diameter \( d \):
\[ A = \frac{\pi d^2}{4} \]
- Therefore, we can write:
\[ A_A = \frac{\pi d_A^2}{4} \quad \text{and} \quad A_B = \frac{\pi d_B^2}{4} \]

4. **Substitute Areas into the Equation of Continuity:**
- Substitute \( A_A \) and \( A_B \) into the equation of continuity:
\[ \frac{\pi d_A^2}{4} \cdot v_A = \frac{\pi d_B^2}{4} \cdot v_B \]
- The \( \frac{\pi}{4} \) cancels out:
\[ d_A^2 v_A = d_B^2 v_B \]

5. **Substitute the Velocity Ratio:**
- Substitute \( v_A = 4v_B \) into the equation:
\[ d_A^2 (4v_B) = d_B^2 v_B \]
- Cancel \( v_B \) (assuming \( v_B \neq 0 \)):
\[ 4 d_A^2 = d_B^2 \]

6. **Solve for the Diameter Ratio:**
- Rearranging gives:
\[ \frac{d_B^2}{d_A^2} = 4 \]
- Taking the square root of both sides:
\[ \frac{d_B}{d_A} = 2 \]
- Therefore, the ratio of the diameters is:
\[ \frac{d_A}{d_B} = \frac{1}{2} \]

### Final Answer:
The ratio of the diameters of the pipe at points A and B is:
\[ \frac{d_A}{d_B} = \frac{1}{2} \]