Two bodies of masses M and 27 M are allowed to fall on a viscous liquid simultaneously. The respective terminal velocities of bodies are V and kV. The value of k is

Options

A. 5
B. 7
C. 8
D. 9

Explanation

To solve the problem, we need to find the value of \( k \) in the context of two bodies falling through a viscous liquid, where one body has a mass \( M \) and the other has a mass \( 27M \). The terminal velocities of these bodies are \( V \) and \( kV \) respectively.

### Step-by-Step Solution:

1. **Understand the Terminal Velocity Formula**:
The terminal velocity \( V_t \) of an object falling through a viscous fluid is given by the formula:
\[ V_t = \frac{2}{9} \frac{r^2 (\rho - \sigma) g}{\eta} \]
where:
- \( r \) = radius of the object,
- \( \rho \) = density of the object,
- \( \sigma \) = density of the fluid,
- \( g \) = acceleration due to gravity,
- \( \eta \) = viscosity of the fluid.

2. **Relate Mass and Volume**:
The density \( \rho \) of the bodies can be expressed as:
\[ \rho = \frac{M}{V} \]
where \( V \) is the volume of the body. The volume of a sphere is given by \( V = \frac{4}{3} \pi r^3 \). Thus, we can express the density in terms of radius:
\[ \rho = \frac{M}{\frac{4}{3} \pi r^3} \implies \rho \propto \frac{M}{r^3} \]

3. **Set Up the Ratios**:
For the two bodies, let:
- Body 1 (mass \( M \)): terminal velocity \( V \)
- Body 2 (mass \( 27M \)): terminal velocity \( kV \)

Using the terminal velocity formula, we can write:
\[ V \propto \frac{r_1^2 (\rho_1 - \sigma) g}{\eta} \]
\[ kV \propto \frac{r_2^2 (\rho_2 - \sigma) g}{\eta} \]

4. **Density Relation**:
Since the density of the fluid \( \sigma \) is constant, we can relate the densities of the two bodies:
\[ \rho_1 = \frac{M}{\frac{4}{3} \pi r_1^3}, \quad \rho_2 = \frac{27M}{\frac{4}{3} \pi r_2^3} \]
Thus,
\[ \frac{\rho_1}{\rho_2} = \frac{M / \frac{4}{3} \pi r_1^3}{27M / \frac{4}{3} \pi r_2^3} = \frac{r_2^3}{27 r_1^3} \]

5. **Volume Ratio**:
Since the densities are equal at terminal velocity, we can set up the ratio:
\[ \frac{r_1^3}{r_2^3} = \frac{1}{27} \implies \frac{r_1}{r_2} = \frac{1}{3} \]

6. **Terminal Velocity Ratio**:
Now, substituting this ratio into the terminal velocity equations:
\[ \frac{V}{kV} = \frac{r_1^2}{r_2^2} = \left(\frac{1}{3}\right)^2 = \frac{1}{9} \]
Thus, we have:
\[ \frac{1}{k} = \frac{1}{9} \implies k = 9 \]

### Conclusion:
The value of \( k \) is \( 9 \).