A sphere of brass released in a long liquid column attains a terminal speed `v_(0)`. If the terminal speed is attained by a sphere of marble of the same radius and released in the same liquid is `nv_(0)`, then the value of `n` will be (Given: The specific gravities of brass, marble and liquid are `8.5, 2.5` and `0.8`, respectively)

Options

A. `5/17`
B. `17/77`
C. `1/31`
D. `17/5`

Explanation

To find the value of \( n \) in the problem, we need to analyze the terminal velocities of the two spheres (brass and marble) in the liquid. We will use the formula for terminal velocity and the specific gravities provided.

### Step 1: Understand the formula for terminal velocity
The terminal velocity \( v_t \) of a sphere falling through a fluid is given by the equation:
\[ v_t = \frac{2}{9} \frac{r^2 (\rho_{object} - \rho_{liquid}) g}{\eta} \]
where:
- \( r \) is the radius of the sphere,
- \( \rho_{object} \) is the density of the object,
- \( \rho_{liquid} \) is the density of the liquid,
- \( g \) is the acceleration due to gravity,
- \( \eta \) is the viscosity of the liquid.

### Step 2: Identify the specific gravities and convert them to densities
Given:
- Specific gravity of brass \( SG_{brass} = 8.5 \)
- Specific gravity of marble \( SG_{marble} = 2.5 \)
- Specific gravity of liquid \( SG_{liquid} = 0.8 \)

The density of a substance can be calculated using its specific gravity:
\[ \rho = SG \times \rho_{water} \]
Assuming the density of water \( \rho_{water} \) is \( 1000 \, \text{kg/m}^3 \):
- Density of brass \( \rho_{brass} = 8.5 \times 1000 = 8500 \, \text{kg/m}^3 \)
- Density of marble \( \rho_{marble} = 2.5 \times 1000 = 2500 \, \text{kg/m}^3 \)
- Density of liquid \( \rho_{liquid} = 0.8 \times 1000 = 800 \, \text{kg/m}^3 \)

### Step 3: Write the terminal velocity equations for both spheres
For the brass sphere:
\[ v_0 = \frac{2}{9} \frac{r^2 (8500 - 800) g}{\eta} \]

For the marble sphere:
\[ v_m = \frac{2}{9} \frac{r^2 (2500 - 800) g}{\eta} \]

### Step 4: Find the ratio of the terminal velocities
We know that \( v_m = n v_0 \). Therefore, we can write:
\[ \frac{v_m}{v_0} = n \]
Substituting the expressions for \( v_m \) and \( v_0 \):
\[ n = \frac{\frac{2}{9} \frac{r^2 (2500 - 800) g}{\eta}}{\frac{2}{9} \frac{r^2 (8500 - 800) g}{\eta}} \]

### Step 5: Simplify the equation
The terms \( \frac{2}{9} \), \( r^2 \), \( g \), and \( \eta \) cancel out:
\[ n = \frac{(2500 - 800)}{(8500 - 800)} = \frac{1700}{7700} = \frac{17}{77} \]

### Conclusion
Thus, the value of \( n \) is:
\[ n = \frac{17}{77} \]