An ideal fluid flows through a pipe of circular cross-section made of two sections with diameters `2.5 cm` and `3.75 cm`. The ratio of the velocities in the two pipes is

Options

A. `9:4`
B. `3:2`
C. `sqrt(3):sqrt(2)`
D. `sqrt(2):sqrt(3)`

Explanation

To find the ratio of the velocities in the two sections of the pipe, we can use the principle of conservation of mass, which is represented by the equation of continuity for fluid flow. The equation states that the mass flow rate must be constant throughout the pipe.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Diameter of section 1 (d1) = 2.5 cm
- Diameter of section 2 (d2) = 3.75 cm

2. **Calculate the Cross-Sectional Areas:**
The cross-sectional area (A) of a circular pipe can be calculated using the formula:
\[ A = \frac{\pi d^2}{4} \]
- For section 1 (A1):
\[ A_1 = \frac{\pi (2.5)^2}{4} = \frac{\pi \cdot 6.25}{4} = \frac{6.25\pi}{4} \]
- For section 2 (A2):
\[ A_2 = \frac{\pi (3.75)^2}{4} = \frac{\pi \cdot 14.0625}{4} = \frac{14.0625\pi}{4} \]

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

4. **Substituting the Areas:**
Substitute the expressions for A1 and A2 into the ratio:
\[ \frac{V_1}{V_2} = \frac{\frac{14.0625\pi}{4}}{\frac{6.25\pi}{4}} \]
The \(\pi\) and \(\frac{1}{4}\) cancel out:
\[ \frac{V_1}{V_2} = \frac{14.0625}{6.25} \]

5. **Calculating the Ratio:**
Simplifying the fraction:
\[ \frac{14.0625}{6.25} = \frac{14.0625 \div 6.25}{6.25 \div 6.25} = \frac{2.25}{1} = 2.25 \]
To express this as a fraction:
\[ 2.25 = \frac{9}{4} \]

6. **Final Result:**
Therefore, the ratio of the velocities \(V_1 : V_2\) is:
\[ \frac{V_1}{V_2} = \frac{9}{4} \]

### Final Answer:
The ratio of the velocities in the two pipes is \(9 : 4\).

---