A small sphere of radius R is held against the inner surface of a larger sphere of radius 6R. The mass of large and small spheres are 4M and M respectively. This arrangement is placed on a horizontal table. There is no friction between any surface of contact. The small sphere is now released. The x coordinate of the centre of the larger sphere when the smaller sphere reaches the other extreme position, is found to be `(L + nR)` , find n.

Explanation

To solve the problem step by step, we will analyze the motion of the small sphere and the resulting motion of the larger sphere.

### Step 1: Understand the initial setup
We have a small sphere of radius \( R \) and mass \( M \) held against the inner surface of a larger sphere of radius \( 6R \) and mass \( 4M \). The center of the larger sphere is at point \( O \) on the x-axis, and the small sphere is initially at a distance of \( 6R - R = 5R \) from the center of the larger sphere.

### Step 2: Determine the initial position of the center of mass
The initial position of the center of mass \( x_i \) can be calculated using the formula for the center of mass of a system:
\[ x_i = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2} \]
Where:
- \( m_1 = 4M \) (mass of the larger sphere)
- \( m_2 = M \) (mass of the smaller sphere)
- \( x_1 = L \) (x-coordinate of the center of the larger sphere)
- \( x_2 = L + 5R \) (x-coordinate of the small sphere)

Substituting these values:
\[ x_i = \frac{(4M)(L) + (M)(L + 5R)}{4M + M} \]
\[ x_i = \frac{4ML + ML + 5MR}{5M} = \frac{5ML + 5MR}{5M} = L + R \]

### Step 3: Determine the final position of the center of mass
When the small sphere is released, it will roll to the other extreme position against the inner surface of the larger sphere. The final position of the small sphere will be at a distance of \( 6R - R = 5R \) from the center of the larger sphere, but now it will be on the opposite side.

The final position of the small sphere can be represented as:
\[ x_2' = L - 5R \]
Now, we can find the final position of the center of mass \( x_f \):
\[ x_f = \frac{(4M)(L) + (M)(L - 5R)}{4M + M} \]
\[ x_f = \frac{(4ML) + (ML - 5MR)}{5M} = \frac{4ML + ML - 5MR}{5M} = \frac{5ML - 5MR}{5M} = L - R \]

### Step 4: Set the initial and final center of mass equal
Since there are no external forces acting on the system, the center of mass remains constant:
\[ x_i = x_f \]
Substituting the values we found:
\[ L + R = L - R \]

### Step 5: Solve for \( n \)
From the equation \( x_f = L + nR \), we can equate:
\[ L - R = L + nR \]
Subtracting \( L \) from both sides gives:
\[ -R = nR \]
Dividing both sides by \( R \) (assuming \( R \neq 0 \)):
\[ -1 = n \]

### Conclusion
The value of \( n \) is \( -1 \).