A metal ball `B_(1)` (density `3.2g//"cc")` is dropped in water, while another metal ball `B_(2)` (density `6.0g//"cc")` is dropped in a liquid of density `1.6g//"cc"`. If both the balls have the same diameter and attain the same terminal velocity, the ratio of viscosity of water to that of the liquid is

Options

A. 2
B. 0.5
C. 4
D. indeterminate due to insufficient data

Explanation

To solve the problem, we will use the concept of terminal velocity and the relationship between the viscosity of fluids and the forces acting on the falling balls.

### Step-by-Step Solution:

1. **Understanding the Problem**:
We have two metal balls, B1 and B2, with different densities, dropped in different fluids. We need to find the ratio of the viscosity of water to that of the liquid in which B2 is dropped.

2. **Given Data**:
- Density of ball B1, \( \sigma_1 = 3.2 \, \text{g/cm}^3 \)
- Density of ball B2, \( \sigma_2 = 6.0 \, \text{g/cm}^3 \)
- Density of water, \( \rho_w = 1.0 \, \text{g/cm}^3 \)
- Density of the liquid, \( \rho_l = 1.6 \, \text{g/cm}^3 \)
- Both balls have the same diameter and attain the same terminal velocity.

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

4. **Setting Up the Equations**:
For ball B1 (in water):
\[ V_{t1} = \frac{2}{9} \frac{r^2 g (\sigma_1 - \rho_w)}{\eta_w} \]
For ball B2 (in liquid):
\[ V_{t2} = \frac{2}{9} \frac{r^2 g (\sigma_2 - \rho_l)}{\eta_l} \]

5. **Equating the Terminal Velocities**:
Since both balls attain the same terminal velocity:
\[ \frac{2}{9} \frac{r^2 g (\sigma_1 - \rho_w)}{\eta_w} = \frac{2}{9} \frac{r^2 g (\sigma_2 - \rho_l)}{\eta_l} \]
We can cancel out common terms:
\[ \frac{(\sigma_1 - \rho_w)}{\eta_w} = \frac{(\sigma_2 - \rho_l)}{\eta_l} \]

6. **Finding the Ratio of Viscosities**:
Rearranging the equation gives:
\[ \frac{\eta_w}{\eta_l} = \frac{\sigma_1 - \rho_w}{\sigma_2 - \rho_l} \]

7. **Substituting the Values**:
- For ball B1:
\[ \sigma_1 - \rho_w = 3.2 - 1.0 = 2.2 \, \text{g/cm}^3 \]
- For ball B2:
\[ \sigma_2 - \rho_l = 6.0 - 1.6 = 4.4 \, \text{g/cm}^3 \]
Therefore,
\[ \frac{\eta_w}{\eta_l} = \frac{2.2}{4.4} = \frac{1}{2} \]

8. **Final Result**:
The ratio of the viscosity of water to that of the liquid is:
\[ \frac{\eta_w}{\eta_l} = \frac{1}{2} \]