Four identical solid spheres each of mass `M` and radius `R` are fixed at four corners of a light square frame of side length `4R` such that centres of spheres coincides with corners of square, moment of inertia of `4` spheres about an axis passing through any side of square is

Options

A. `(21 MR^2)/(5)`
B. `(42 MR^2)/(5)`
C. `(84 MR^2)/(5)`
D. `(168 MR^2)/(5)`

Answer

To solve the problem of finding the moment of inertia of four identical solid spheres, each of mass \( M \) and radius \( R \), fixed at the corners of a square frame of side length \( 4R \), we will follow these steps:

### Step 1: Understand the Configuration
We have four identical solid spheres located at the corners of a square with side length \( 4R \). The centers of the spheres coincide with the corners of the square.

### Step 2: Identify the Axis of Rotation
We need to calculate the moment of inertia about an axis that passes through one side of the square. Let's denote the corners of the square as A, B, C, and D.

### Step 3: Calculate the Moment of Inertia for Spheres A and B
The spheres at corners A and B lie on the axis of rotation. The moment of inertia \( I \) for a solid sphere about its own axis is given by:

\[ I = \frac{2}{5} M R^2 \]

Since both spheres A and B are on the axis, their contributions to the moment of inertia are:

\[ I_A = I_B = \frac{2}{5} M R^2 \]

Thus, the total moment of inertia for spheres A and B is:

\[ I_{AB} = I_A + I_B = 2 \times \frac{2}{5} M R^2 = \frac{4}{5} M R^2 \]

### Step 4: Calculate the Moment of Inertia for Spheres C and D
Spheres C and D are not on the axis of rotation. We will use the parallel axis theorem to find their contributions. The distance from the axis to the centers of spheres C and D is \( 2R \) (half the side length of the square).

Using the parallel axis theorem:

\[ I = I_{cm} + Md^2 \]

where \( I_{cm} = \frac{2}{5} M R^2 \) (moment of inertia about their own center) and \( d = 2R \).

Thus, for spheres C and D:

\[ I_C = \frac{2}{5} M R^2 + M(2R)^2 = \frac{2}{5} M R^2 + 4M R^2 = \frac{2}{5} M R^2 + \frac{20}{5} M R^2 = \frac{22}{5} M R^2 \]

The same calculation applies to sphere D:

\[ I_D = \frac{22}{5} M R^2 \]

### Step 5: Total Moment of Inertia
Now we can find the total moment of inertia \( I \) by summing the contributions from all four spheres:

\[ I = I_{AB} + I_C + I_D \]

Substituting the values we found:

\[ I = \frac{4}{5} M R^2 + \frac{22}{5} M R^2 + \frac{22}{5} M R^2 \]

Calculating this gives:

\[ I = \frac{4}{5} M R^2 + \frac{44}{5} M R^2 = \frac{48}{5} M R^2 \]

### Final Answer
Thus, the moment of inertia of the four spheres about an axis passing through any side of the square is:

\[ \boxed{\frac{48}{5} M R^2} \]