An ideal fluid flows through a pipe of circular cross section made of two section with diamters 2.5 cm and 9.76 cm. The ratio of the velocities in the two pipes is

Options

A. `9:4`
B. `3:2`
C. `sqrt3:sqrt2`
D. `sqrt2:sqrt3`

Explanation

To solve the problem of finding the ratio of velocities in two sections of a pipe with different diameters, we can use the principle of conservation of mass for incompressible fluids, which states that the product of the cross-sectional area and the velocity of flow must remain constant throughout the pipe. This is known as the equation of continuity.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Diameter of the first section (D1) = 2.5 cm
- Diameter of the second section (D2) = 9.76 cm

2. **Calculate the Cross-Sectional Areas:**
- The cross-sectional area (A) of a circular pipe is given by the formula:
\[ A = \pi \left(\frac{D}{2}\right)^2 \]
- For the first section (A1):
\[ A1 = \pi \left(\frac{D1}{2}\right)^2 = \pi \left(\frac{2.5}{2}\right)^2 = \pi \left(1.25\right)^2 = \pi \times 1.5625 \]
- For the second section (A2):
\[ A2 = \pi \left(\frac{D2}{2}\right)^2 = \pi \left(\frac{9.76}{2}\right)^2 = \pi \left(4.88\right)^2 = \pi \times 23.8144 \]

3. **Apply the Equation of Continuity:**
- According to the equation of continuity:
\[ A1 \cdot V1 = A2 \cdot V2 \]
- Rearranging gives us:
\[ \frac{V1}{V2} = \frac{A2}{A1} \]

4. **Substituting the Areas:**
- Substitute the areas calculated:
\[ \frac{V1}{V2} = \frac{A2}{A1} = \frac{\pi \times 23.8144}{\pi \times 1.5625} \]
- The \(\pi\) cancels out:
\[ \frac{V1}{V2} = \frac{23.8144}{1.5625} \]

5. **Calculate the Ratio:**
- Performing the division:
\[ \frac{V1}{V2} = 15.21 \]

6. **Conclusion:**
- Therefore, the ratio of the velocities \(V1\) to \(V2\) is:
\[ V1 : V2 = 15.21 : 1 \]