An ideal fluid is flowing in a pipe in streamline flow. Pipe has maximum and minimum diameter of 6.4 cm and 4.8 cm respectively. Find out the ratio of minimum to maximum velocity.

Options

A. `9/16`
B. `8/17`
C. `5/12`
D. `3/4`

Explanation

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 the fluid must remain constant along a streamline.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Maximum diameter of the pipe, \( D_1 = 6.4 \, \text{cm} \)
- Minimum diameter of the pipe, \( D_2 = 4.8 \, \text{cm} \)

2. **Convert Diameters to Meters:**
- \( D_1 = 0.064 \, \text{m} \)
- \( D_2 = 0.048 \, \text{m} \)

3. **Calculate the Cross-Sectional Areas:**
- The area \( A \) of a circular cross-section is given by the formula:
\[ A = \frac{\pi D^2}{4} \]
- For the maximum diameter \( D_1 \):
\[ A_1 = \frac{\pi (0.064)^2}{4} = \frac{\pi \times 0.004096}{4} = \frac{0.004096\pi}{4} \approx 0.003219 \, \text{m}^2 \]
- For the minimum diameter \( D_2 \):
\[ A_2 = \frac{\pi (0.048)^2}{4} = \frac{\pi \times 0.002304}{4} = \frac{0.002304\pi}{4} \approx 0.001823 \, \text{m}^2 \]

4. **Apply the Continuity Equation:**
- According to the continuity equation:
\[ A_1 V_1 = A_2 V_2 \]
- Rearranging gives:
\[ \frac{V_2}{V_1} = \frac{A_1}{A_2} \]
- Therefore:
\[ V_1 = \frac{A_2}{A_1} V_2 \]

5. **Substituting the Areas:**
- We can express the ratio of velocities in terms of the diameters:
\[ \frac{V_2}{V_1} = \frac{A_1}{A_2} = \frac{\frac{\pi D_1^2}{4}}{\frac{\pi D_2^2}{4}} = \frac{D_1^2}{D_2^2} \]

6. **Calculating the Ratio:**
- Substitute the diameters:
\[ \frac{V_2}{V_1} = \left(\frac{D_1}{D_2}\right)^2 = \left(\frac{6.4}{4.8}\right)^2 \]
- Simplifying:
\[ \frac{6.4}{4.8} = \frac{64}{48} = \frac{4}{3} \]
- Therefore:
\[ \left(\frac{4}{3}\right)^2 = \frac{16}{9} \]

7. **Finding the Ratio of Minimum to Maximum Velocity:**
- Thus, the ratio of minimum velocity \( V_2 \) to maximum velocity \( V_1 \) is:
\[ \frac{V_2}{V_1} = \frac{9}{16} \]

### Final Answer:
The ratio of minimum to maximum velocity is \( \frac{9}{16} \).

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