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 solve the problem of finding the ratio of the velocities in two sections of a pipe with different diameters, we can use the principle of continuity for an ideal fluid. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the Problem
We are given two sections of a pipe with diameters:
- \( d_1 = 2.5 \, \text{cm} \)
- \( d_2 = 3.75 \, \text{cm} \)

We need to find the ratio of the velocities \( V_1 \) and \( V_2 \) in these two sections.

### Step 2: Use the Equation of Continuity
According to the equation of continuity for an incompressible fluid, the product of the cross-sectional area \( A \) and the velocity \( V \) at any two points in a flow must be constant:
\[ A_1 V_1 = A_2 V_2 \]

### Step 3: Calculate the Areas
The area \( A \) of a circular cross-section is given by:
\[ A = \pi r^2 \]
Where \( r \) is the radius of the pipe. The radius can be calculated from the diameter:
\[ r = \frac{d}{2} \]
Thus, the areas for the two sections are:
\[ A_1 = \pi \left( \frac{d_1}{2} \right)^2 = \frac{\pi d_1^2}{4} \]
\[ A_2 = \pi \left( \frac{d_2}{2} \right)^2 = \frac{\pi d_2^2}{4} \]

### Step 4: Set Up the Equation
Substituting the areas into the continuity equation gives:
\[ \frac{\pi d_1^2}{4} V_1 = \frac{\pi d_2^2}{4} V_2 \]
We can simplify this by canceling \( \pi \) and \( 4 \) from both sides:
\[ d_1^2 V_1 = d_2^2 V_2 \]

### Step 5: Solve for the Velocity Ratio
Rearranging the equation to find the ratio \( \frac{V_1}{V_2} \):
\[ \frac{V_1}{V_2} = \frac{d_2^2}{d_1^2} \]

### Step 6: Substitute the Values
Now, substituting the given diameters:
\[ d_1 = 2.5 \, \text{cm}, \quad d_2 = 3.75 \, \text{cm} \]
Calculating the squares:
\[ \frac{V_1}{V_2} = \frac{(3.75)^2}{(2.5)^2} = \frac{14.0625}{6.25} \]

### Step 7: Simplify the Ratio
Calculating the ratio:
\[ \frac{V_1}{V_2} = \frac{14.0625}{6.25} = 2.25 \]
This can also be expressed as:
\[ \frac{V_1}{V_2} = \frac{9}{4} \]

### Final Answer
Thus, the ratio of the velocities in the two sections of the pipe is:
\[ \frac{V_1}{V_2} = \frac{9}{4} \]