water is flowing through a tube of non-uniform cross-section. If the radii of the tube at the entrance and the exit are in the ratio `3:2` then the ratio of the velocites of flow of watern at the entrance and the exit is

Options

A. `1:1`
B. `4:9`
C. `9:4`
D. `8:27`

Answer

To solve the problem, we will use the principle of continuity for fluid flow, which states that the mass flow rate must remain constant in a tube of varying cross-section. This means that the product of the cross-sectional area and the velocity of the fluid must be the same at any two points along the tube.

### Step-by-Step Solution:

1. **Identify the Given Ratios:**
- The ratio of the radii at the entrance (R1) and the exit (R2) of the tube is given as 3:2.
- Therefore, we can express this as:
\[ \frac{R1}{R2} = \frac{3}{2} \]

2. **Calculate the Areas:**
- The area (A) of a circular cross-section is given by the formula:
\[ A = \pi R^2 \]
- Thus, the area at the entrance (A1) and the exit (A2) can be expressed as:
\[ A1 = \pi R1^2 \quad \text{and} \quad A2 = \pi R2^2 \]

3. **Apply the Continuity Equation:**
- According to the continuity equation:
\[ A1 \cdot V1 = A2 \cdot V2 \]
- Substituting the areas we calculated:
\[ \pi R1^2 \cdot V1 = \pi R2^2 \cdot V2 \]
- The \(\pi\) cancels out:
\[ R1^2 \cdot V1 = R2^2 \cdot V2 \]

4. **Rearranging for Velocity Ratio:**
- Rearranging the equation gives us:
\[ \frac{V1}{V2} = \frac{R2^2}{R1^2} \]

5. **Substituting the Radius Ratio:**
- We know \(\frac{R1}{R2} = \frac{3}{2}\), so:
\[ \frac{R2}{R1} = \frac{2}{3} \]
- Squaring both sides:
\[ \left(\frac{R2}{R1}\right)^2 = \left(\frac{2}{3}\right)^2 = \frac{4}{9} \]

6. **Final Velocity Ratio:**
- Substituting this back into the velocity ratio:
\[ \frac{V1}{V2} = \frac{4}{9} \]
- Therefore, the ratio of the velocities of flow of water at the entrance and the exit is:
\[ V1 : V2 = 4 : 9 \]

### Final Answer:
The ratio of the velocities of flow of water at the entrance and the exit is \(4 : 9\).