Two spheres each of mass `M` and radius `R//2` are connected at their centres with a mass less rod of length `2 R`. What will be the moment of inertia of the system about an axis passing through the centre of one of the sphere and perpendicular to the rod ?

Options

A. `21 MR^2//5`
B. `2 MR^2//5`
C. `5 MR^2//2`
D. `5 MR^2//21`

Explanation

To find the moment of inertia of the system about an axis passing through the center of one of the spheres and perpendicular to the rod, we can follow these steps:

### Step 1: Identify the components of the system
We have two spheres, each with mass \( M \) and radius \( R/2 \). They are connected by a massless rod of length \( 2R \). The axis of rotation is through the center of one sphere.

### Step 2: Calculate the moment of inertia of the first sphere
The moment of inertia \( I_1 \) of a solid sphere about its own center is given by the formula:
\[ I_{\text{sphere}} = \frac{2}{5} M R^2 \]
For the first sphere, substituting \( R = \frac{R}{2} \):
\[ I_1 = \frac{2}{5} M \left(\frac{R}{2}\right)^2 = \frac{2}{5} M \frac{R^2}{4} = \frac{2M R^2}{20} = \frac{M R^2}{10} \]

### Step 3: Calculate the moment of inertia of the second sphere using the parallel axis theorem
The second sphere is at a distance of \( 2R \) from the axis of rotation. We first calculate its moment of inertia about its own center:
\[ I_{\text{sphere}} = \frac{2}{5} M R^2 \]
For the second sphere, substituting \( R = \frac{R}{2} \):
\[ I_2 = \frac{2}{5} M \left(\frac{R}{2}\right)^2 = \frac{M R^2}{10} \]
Now, we apply the parallel axis theorem to find the moment of inertia about the desired axis:
\[ I_2' = I_2 + M d^2 \]
where \( d = 2R \) (the distance between the centers of the two spheres):
\[ I_2' = \frac{M R^2}{10} + M (2R)^2 = \frac{M R^2}{10} + M \cdot 4R^2 = \frac{M R^2}{10} + 4M R^2 \]
Combining the terms:
\[ I_2' = \frac{M R^2}{10} + \frac{40M R^2}{10} = \frac{41M R^2}{10} \]

### Step 4: Calculate the total moment of inertia of the system
Now, we can find the total moment of inertia \( I \) of the system:
\[ I = I_1 + I_2' = \frac{M R^2}{10} + \frac{41M R^2}{10} = \frac{42M R^2}{10} = \frac{21M R^2}{5} \]

### Final Answer
The moment of inertia of the system about the specified axis is:
\[ I = \frac{21M R^2}{5} \]