Two lead balls of masses `m` and `5m` having radii `R` and `2R` are separated by `12R`. If they attract each other by gravitational force, the distance covered by small sphere before they touch each other is

Options

A. `10R`
B. `7.5R`
C. `9R`
D. `2.5 R`

Explanation

To solve the problem, we need to find the distance covered by the smaller lead ball (mass `m` and radius `R`) before they touch each other. The larger lead ball has a mass of `5m` and a radius of `2R`, and they are initially separated by a distance of `12R`.

### Step-by-Step Solution:

1. **Identify the Initial Setup:**
- Mass of the smaller ball, \( m_1 = m \)
- Radius of the smaller ball, \( R_1 = R \)
- Mass of the larger ball, \( m_2 = 5m \)
- Radius of the larger ball, \( R_2 = 2R \)
- Initial separation between the centers of the balls, \( d = 12R \)

2. **Calculate the Distance Between the Centers When They Touch:**
- When the two balls touch, the distance between their centers will be the sum of their radii:
\[ d_{\text{touch}} = R_1 + R_2 = R + 2R = 3R \]

3. **Determine the Final Distance Between the Centers:**
- The initial distance between the centers is \( 12R \). When they touch, the distance will be \( 3R \).
- The distance that needs to be covered for them to touch is:
\[ \Delta d = d - d_{\text{touch}} = 12R - 3R = 9R \]

4. **Calculate the Center of Mass:**
- The center of mass (CM) of the system can be calculated using the formula:
\[ x_{\text{CM}} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2} \]
- Let’s place the smaller ball at the origin (0, 0) and the larger ball at (12R, 0):
\[ x_{\text{CM}} = \frac{m \cdot 0 + 5m \cdot 12R}{m + 5m} = \frac{60mR}{6m} = 10R \]

5. **Determine the Distance Covered by the Smaller Ball:**
- Let the distance covered by the smaller ball before they touch be \( x \).
- The distance covered by the larger ball will be \( 9R - x \) since the total distance to be covered is \( 9R \).
- The ratio of the distances covered by the two balls is inversely proportional to their masses:
\[ \frac{x}{9R - x} = \frac{5m}{m} = 5 \]
- Cross-multiplying gives:
\[ x = 5(9R - x) \]
\[ x = 45R - 5x \]
\[ 6x = 45R \]
\[ x = \frac{45R}{6} = 7.5R \]

### Conclusion:
The distance covered by the smaller sphere before they touch each other is \( 7.5R \).