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

Answer

To solve the problem, we need to determine the ratio of the terminal velocities \( v_1 \) and \( v_2 \) of a solid sphere and smaller spheres, respectively.

### Step-by-Step Solution:

1. **Understanding Terminal Velocity**:
The terminal velocity \( v \) of a sphere falling through a viscous fluid is given by the formula:
\[ v = \frac{2}{9} \frac{R^2 g (\rho_s - \rho_f)}{\eta} \]
where:
- \( R \) is the radius of the sphere,
- \( g \) is the acceleration due to gravity,
- \( \rho_s \) is the density of the sphere,
- \( \rho_f \) is the density of the fluid,
- \( \eta \) is the coefficient of viscosity.

2. **Terminal Velocity of the Original Sphere**:
For the original sphere of radius \( R \), the terminal velocity \( v_1 \) can be expressed as:
\[ v_1 = \frac{2}{9} \frac{R^2 g (\rho_s - \rho_f)}{\eta} \]

3. **Breaking the Sphere into Smaller Spheres**:
The original sphere is broken into 27 identical smaller spheres. The volume of the original sphere is:
\[ V = \frac{4}{3} \pi R^3 \]
The volume of each smaller sphere (let's denote the radius of the smaller sphere as \( r \)) is:
\[ V_{small} = \frac{4}{3} \pi r^3 \]
Since the total volume remains the same, we have:
\[ \frac{4}{3} \pi R^3 = 27 \left(\frac{4}{3} \pi r^3\right) \]
Simplifying this gives:
\[ R^3 = 27 r^3 \implies r^3 = \frac{R^3}{27} \implies r = \frac{R}{3} \]

4. **Terminal Velocity of the Smaller Spheres**:
Now, we can find the terminal velocity \( v_2 \) of one of the smaller spheres with radius \( r = \frac{R}{3} \):
\[ v_2 = \frac{2}{9} \frac{r^2 g (\rho_s - \rho_f)}{\eta} = \frac{2}{9} \frac{\left(\frac{R}{3}\right)^2 g (\rho_s - \rho_f)}{\eta} \]
Simplifying this gives:
\[ v_2 = \frac{2}{9} \frac{\frac{R^2}{9} g (\rho_s - \rho_f)}{\eta} = \frac{2}{81} \frac{R^2 g (\rho_s - \rho_f)}{\eta} \]

5. **Finding the Ratio \( \frac{v_1}{v_2} \)**:
Now we can find the ratio of the terminal velocities:
\[ \frac{v_1}{v_2} = \frac{\frac{2}{9} \frac{R^2 g (\rho_s - \rho_f)}{\eta}}{\frac{2}{81} \frac{R^2 g (\rho_s - \rho_f)}{\eta}} = \frac{\frac{2}{9}}{\frac{2}{81}} = \frac{81}{9} = 9 \]

### Final Answer:
Thus, the ratio \( \frac{v_1}{v_2} \) is \( 9:1 \).