Two solid spheres of same metal but of mass M and 8M fall simultaneously on a viscous liquid and their terminal velocities are v and nv, then value of n is

Options

A. 16
B. 8
C. 4
D. 2

Explanation

To solve the problem, we need to determine the relationship between the terminal velocities of two solid spheres of the same metal but different masses. The masses are \( M \) and \( 8M \), and their terminal velocities are \( v \) and \( nv \) respectively.

### Step-by-step Solution:

1. **Understanding the Relationship between Mass and Volume**:
- The mass of a sphere is given by the formula:
\[ M = \rho V \]
where \( \rho \) is the density and \( V \) is the volume.
- For a sphere, the volume \( V \) is:
\[ V = \frac{4}{3} \pi r^3 \]
- Therefore, the mass can also be expressed as:
\[ M = \rho \left( \frac{4}{3} \pi r^3 \right) \]

2. **Finding the Radius of the Spheres**:
- Let the radius of the first sphere (mass \( M \)) be \( r_1 \) and the radius of the second sphere (mass \( 8M \)) be \( r_2 \).
- From the mass relationship:
\[ M_1 = \rho \left( \frac{4}{3} \pi r_1^3 \right) \quad \text{and} \quad M_2 = 8M = \rho \left( \frac{4}{3} \pi r_2^3 \right) \]
- Dividing the two equations:
\[ \frac{M_1}{M_2} = \frac{r_1^3}{r_2^3} \implies \frac{M}{8M} = \frac{r_1^3}{r_2^3} \implies \frac{1}{8} = \left( \frac{r_1}{r_2} \right)^3 \]
- Taking the cube root:
\[ \frac{r_1}{r_2} = \frac{1}{2} \implies r_1 = \frac{1}{2} r_2 \]

3. **Using the Terminal Velocity Formula**:
- The terminal velocity \( v_t \) of a sphere falling through a viscous liquid is given by:
\[ v_t = \frac{2}{9} \frac{r^2 (\rho - \sigma) g}{\eta} \]
where \( \sigma \) is the density of the liquid, \( g \) is the acceleration due to gravity, and \( \eta \) is the viscosity of the liquid.
- Since \( \rho \), \( \sigma \), \( g \), and \( \eta \) are constants, the terminal velocity is directly proportional to the square of the radius:
\[ v_t \propto r^2 \]

4. **Finding the Ratio of Terminal Velocities**:
- For the first sphere:
\[ v_1 \propto r_1^2 \]
- For the second sphere:
\[ v_2 \propto r_2^2 \]
- Thus, the ratio of their terminal velocities is:
\[ \frac{v_1}{v_2} = \frac{r_1^2}{r_2^2} \]

5. **Substituting the Radius Ratio**:
- We know \( \frac{r_1}{r_2} = \frac{1}{2} \), so:
\[ \frac{v_1}{v_2} = \left( \frac{1}{2} \right)^2 = \frac{1}{4} \]
- Therefore, if \( v_1 = v \) and \( v_2 = nv \):
\[ \frac{v}{nv} = \frac{1}{4} \implies n = 4 \]

### Final Answer:
The value of \( n \) is \( 4 \).