Two rain drops reach the earth with different terminal velocities having ratio `9:4` . Then , the ratio of their volumes is

Options

A. `3:2`
B. `4:9`
C. `9:4`
D. `27:8`

Answer

To solve the problem, we need to find the ratio of the volumes of two raindrops given the ratio of their terminal velocities.

### Step-by-Step Solution:

1. **Understand the relationship between terminal velocity and radius**:
The terminal velocity \( V_T \) of a drop is given by the equation:
\[ V_T = \frac{2R^2 g (\rho - \sigma)}{9\eta} \]
where:
- \( R \) is the radius of the drop,
- \( g \) is the acceleration due to gravity,
- \( \rho \) is the density of the liquid,
- \( \sigma \) is the density of the air,
- \( \eta \) is the coefficient of viscosity.

Since all variables except the radius \( R \) are constant for both drops, we can say that the terminal velocity is proportional to the square of the radius:
\[ V_T \propto R^2 \]

2. **Set up the ratio of terminal velocities**:
Given the ratio of terminal velocities of the two raindrops is:
\[ \frac{V_{T1}}{V_{T2}} = \frac{9}{4} \]
Using the proportionality, we can express this in terms of the radii:
\[ \frac{R_1^2}{R_2^2} = \frac{9}{4} \]

3. **Find the ratio of the radii**:
Taking the square root of both sides gives us the ratio of the radii:
\[ \frac{R_1}{R_2} = \frac{3}{2} \]

4. **Calculate the ratio of volumes**:
The volume \( V \) of a sphere is given by:
\[ V = \frac{4}{3} \pi R^3 \]
Therefore, the ratio of the volumes of the two drops is:
\[ \frac{V_1}{V_2} = \frac{\frac{4}{3} \pi R_1^3}{\frac{4}{3} \pi R_2^3} = \frac{R_1^3}{R_2^3} \]
Substituting the ratio of the radii:
\[ \frac{V_1}{V_2} = \left(\frac{3}{2}\right)^3 = \frac{27}{8} \]

5. **Conclusion**:
Therefore, the ratio of the volumes of the two raindrops is:
\[ \frac{V_1}{V_2} = \frac{27}{8} \]

### Final Answer:
The ratio of their volumes is \( 27:8 \).

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