A solid sphere, of radius R acquires a terminal velocity ` v_1 ` when falling (due to gravity) through a viscous fluid having a coefficient of viscosity ` eta ` he sphere is broken into 27 identical solid spheres. If each of these spheres acquires a terminal velocity ` v _2 ` when falling through the same fluid, the ratio ` (v_1//v_2 ) ` equals:

Options

A. `1//27 `
B. `27`
C. 9
D. `1//9`

Explanation

To find the ratio of terminal velocities \( \frac{v_1}{v_2} \) for a solid sphere and the smaller spheres formed from it, we can follow these steps:

### Step-by-Step Solution

1. **Understanding the Problem**:
- A solid sphere of radius \( R \) falls through a viscous fluid and acquires a terminal velocity \( v_1 \).
- This sphere is broken into 27 identical smaller spheres.
- We need to find the terminal velocity \( v_2 \) for each of these smaller spheres and the ratio \( \frac{v_1}{v_2} \).

2. **Volume Conservation**:
- The volume of the original sphere is given by:
\[ V = \frac{4}{3} \pi R^3 \]
- The volume of one of the smaller spheres (let's denote its radius as \( r' \)) is:
\[ V' = \frac{4}{3} \pi (r')^3 \]
- Since the total volume is conserved when the original sphere is broken into 27 smaller spheres, we have:
\[ \frac{4}{3} \pi R^3 = 27 \cdot \frac{4}{3} \pi (r')^3 \]
- Canceling \( \frac{4}{3} \pi \) from both sides gives:
\[ R^3 = 27 (r')^3 \]

3. **Finding the Radius of the Smaller Spheres**:
- Rearranging the equation gives:
\[ (r')^3 = \frac{R^3}{27} \]
- Taking the cube root:
\[ r' = \frac{R}{3} \]

4. **Terminal Velocity Formula**:
- The terminal velocity \( v \) of a sphere falling through a viscous fluid is given by:
\[ v = \frac{2}{9} \frac{g r^2 (\rho - \sigma)}{\eta} \]
- Here, \( g \) is the acceleration due to gravity, \( \rho \) is the density of the fluid, \( \sigma \) is the density of the sphere, and \( \eta \) is the coefficient of viscosity.

5. **Expressing Terminal Velocities**:
- For the original sphere (radius \( R \)):
\[ v_1 = \frac{2}{9} \frac{g R^2 (\rho - \sigma)}{\eta} \]
- For the smaller spheres (radius \( r' = \frac{R}{3} \)):
\[ v_2 = \frac{2}{9} \frac{g (r')^2 (\rho - \sigma)}{\eta} = \frac{2}{9} \frac{g \left(\frac{R}{3}\right)^2 (\rho - \sigma)}{\eta} \]
- Simplifying \( v_2 \):
\[ v_2 = \frac{2}{9} \frac{g \frac{R^2}{9} (\rho - \sigma)}{\eta} = \frac{1}{9} \cdot \frac{2}{9} \frac{g R^2 (\rho - \sigma)}{\eta} = \frac{1}{9} v_1 \]

6. **Finding the Ratio**:
- Now, we can find the ratio \( \frac{v_1}{v_2} \):
\[ \frac{v_1}{v_2} = \frac{v_1}{\frac{1}{9} v_1} = 9 \]

### Final Answer
Thus, the ratio \( \frac{v_1}{v_2} \) is:
\[ \frac{v_1}{v_2} = 9 \]