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. 0.25

Answer

To solve the problem, we will use the concept of terminal velocity and the relationship between the densities of the balls and the fluids they are dropped into.

### Step-by-Step Solution:

1. **Understanding Terminal Velocity**:
The terminal velocity \( V_t \) of a sphere falling through a fluid is given by the equation:
\[ V_t = \frac{2}{9} \frac{R^2 ( \rho_b - \rho_m ) g}{\eta} \]
where:
- \( R \) = radius of the ball
- \( \rho_b \) = density of the ball
- \( \rho_m \) = density of the medium (fluid)
- \( g \) = acceleration due to gravity
- \( \eta \) = coefficient of viscosity of the medium

2. **Setting Up Equations for Each Ball**:
For ball \( B_1 \) (density \( \rho_{b1} = 3.2 \, \text{g/cm}^3 \), in water with \( \rho_{m1} = 1 \, \text{g/cm}^3 \)):
\[ V_{t1} = \frac{2}{9} \frac{R^2 (3.2 - 1) g}{\eta_1} \]
For ball \( B_2 \) (density \( \rho_{b2} = 6.0 \, \text{g/cm}^3 \), in liquid with \( \rho_{m2} = 1.6 \, \text{g/cm}^3 \)):
\[ V_{t2} = \frac{2}{9} \frac{R^2 (6.0 - 1.6) g}{\eta_2} \]

3. **Equating Terminal Velocities**:
Since both balls attain the same terminal velocity (\( V_{t1} = V_{t2} \)):
\[ \frac{2}{9} \frac{R^2 (3.2 - 1) g}{\eta_1} = \frac{2}{9} \frac{R^2 (6.0 - 1.6) g}{\eta_2} \]
The \( \frac{2}{9} R^2 g \) terms cancel out:
\[ \frac{(3.2 - 1)}{\eta_1} = \frac{(6.0 - 1.6)}{\eta_2} \]

4. **Substituting Values**:
Calculate the differences in densities:
\[ 3.2 - 1 = 2.2 \quad \text{and} \quad 6.0 - 1.6 = 4.4 \]
Therefore, we have:
\[ \frac{2.2}{\eta_1} = \frac{4.4}{\eta_2} \]

5. **Finding the Ratio of Viscosities**:
Rearranging gives:
\[ \frac{\eta_1}{\eta_2} = \frac{2.2}{4.4} = \frac{1}{2} \]

6. **Final Result**:
Thus, the ratio of the viscosity of water to that of the liquid is:
\[ \frac{\eta_1}{\eta_2} = 0.5 \]