Two metal spheres are falling through a liquid of density `2xx10^(3)kg//m^(3)` with the same uniform speed. The material density of sphere 1 and sphere 2 are `8xx10^(3)kg//m^(3)` and `11xx10^(3)kg//m^(3)` respectively. The ratio of their radii is :-

Options

A. `(11)/(8)`
B. `sqrt((11)/(8))`
C. `(3)/(2)`
D. `sqrt((3)/(2))`

Answer

To solve the problem of finding the ratio of the radii of two metal spheres falling through a liquid with the same uniform speed, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Condition of Terminal Velocity**:
Since both spheres are falling with the same uniform speed, they are at terminal velocity. The terminal velocity \( v_t \) for a sphere falling through a fluid is given by the formula:
\[ v_t = \frac{2}{9} \frac{r^2 g}{\eta} (\rho_s - \rho_l) \]
where:
- \( r \) = radius of the sphere
- \( g \) = acceleration due to gravity
- \( \eta \) = viscosity of the liquid
- \( \rho_s \) = density of the sphere
- \( \rho_l \) = density of the liquid

2. **Set Up the Equation for Both Spheres**:
Let \( r_1 \) and \( r_2 \) be the radii of sphere 1 and sphere 2, respectively. The densities are given as:
- \( \rho_{s1} = 8 \times 10^3 \, \text{kg/m}^3 \)
- \( \rho_{s2} = 11 \times 10^3 \, \text{kg/m}^3 \)
- \( \rho_l = 2 \times 10^3 \, \text{kg/m}^3 \)

Since both spheres have the same terminal velocity, we can equate their terminal velocity equations:
\[ \frac{2}{9} \frac{r_1^2 g}{\eta} (\rho_{s1} - \rho_l) = \frac{2}{9} \frac{r_2^2 g}{\eta} (\rho_{s2} - \rho_l) \]

3. **Cancel Common Terms**:
The terms \( \frac{2}{9} g \) and \( \eta \) are common on both sides, so they can be canceled out:
\[ r_1^2 (\rho_{s1} - \rho_l) = r_2^2 (\rho_{s2} - \rho_l) \]

4. **Rearranging for the Ratio of Radii**:
Rearranging gives us:
\[ \frac{r_1^2}{r_2^2} = \frac{\rho_{s2} - \rho_l}{\rho_{s1} - \rho_l} \]
Taking the square root of both sides:
\[ \frac{r_1}{r_2} = \sqrt{\frac{\rho_{s2} - \rho_l}{\rho_{s1} - \rho_l}} \]

5. **Substituting the Values**:
Now substitute the values:
- \( \rho_{s2} - \rho_l = 11 \times 10^3 - 2 \times 10^3 = 9 \times 10^3 \)
- \( \rho_{s1} - \rho_l = 8 \times 10^3 - 2 \times 10^3 = 6 \times 10^3 \)

Thus, we have:
\[ \frac{r_1}{r_2} = \sqrt{\frac{9 \times 10^3}{6 \times 10^3}} = \sqrt{\frac{9}{6}} = \sqrt{\frac{3}{2}} \]

6. **Final Ratio**:
Therefore, the ratio of the radii \( r_1 : r_2 \) is:
\[ r_1 : r_2 = \sqrt{\frac{3}{2}} \approx 1.2247 \]

### Conclusion:
The ratio of the radii of the two spheres is \( \sqrt{\frac{3}{2}} \).