The moment of inertia of a hollow cubical box of mass `M` and side `a` about an axis passing through the centres of two opposite faces is equal to.

Options

A. `(5 M a^2)/(3)`
B. `(5 M a^2)/(6)`
C. `(5 M a^2)/(12)`
D. `(5 M a^2)/(18)`

Explanation

To find the moment of inertia of a hollow cubical box of mass \( M \) and side \( a \) about an axis passing through the centers of two opposite faces, we can follow these steps:

### Step 1: Understand the Structure of the Hollow Cube
A hollow cube has mass concentrated on its faces, with the interior being empty. Each face of the cube contributes to the moment of inertia.

### Step 2: Identify the Mass Distribution
Since the cube has six faces and the total mass is \( M \), the mass of each face can be calculated as:
\[ m_{\text{face}} = \frac{M}{6} \]

### Step 3: Calculate the Moment of Inertia for One Face
The moment of inertia \( I \) of a single face about an axis through its center (perpendicular to the face) is given by:
\[ I_{\text{face}} = \frac{1}{12} m_{\text{face}} a^2 \]
Substituting \( m_{\text{face}} \):
\[ I_{\text{face}} = \frac{1}{12} \left(\frac{M}{6}\right) a^2 = \frac{Ma^2}{72} \]

### Step 4: Calculate the Moment of Inertia for the Four Side Faces
The four side faces are symmetrically aligned with the axis. The moment of inertia for each of these faces about the same axis can be calculated using the parallel axis theorem. The distance \( d \) from the center of each face to the axis is \( \frac{a}{2} \). Therefore, the moment of inertia for one side face using the parallel axis theorem is:
\[ I_{\text{side}} = I_{\text{face}} + m_{\text{face}} \left(\frac{a}{2}\right)^2 \]
Substituting the values:
\[ I_{\text{side}} = \frac{Ma^2}{72} + \left(\frac{M}{6}\right) \left(\frac{a^2}{4}\right) = \frac{Ma^2}{72} + \frac{Ma^2}{24} \]
Finding a common denominator (72):
\[ I_{\text{side}} = \frac{Ma^2}{72} + \frac{3Ma^2}{72} = \frac{4Ma^2}{72} = \frac{Ma^2}{18} \]
Since there are four side faces:
\[ I_{\text{sides total}} = 4 \times \frac{Ma^2}{18} = \frac{4Ma^2}{18} = \frac{2Ma^2}{9} \]

### Step 5: Calculate the Moment of Inertia for the Front and Back Faces
For the front and back faces, the moment of inertia about the same axis is:
\[ I_{\text{front}} = I_{\text{back}} = \frac{Ma^2}{72} \]
Thus, the total moment of inertia for the front and back faces is:
\[ I_{\text{front + back}} = 2 \times \frac{Ma^2}{72} = \frac{Ma^2}{36} \]

### Step 6: Combine All Contributions
Now, we can sum up all contributions to find the total moment of inertia \( I_{\text{total}} \):
\[ I_{\text{total}} = I_{\text{sides total}} + I_{\text{front + back}} = \frac{2Ma^2}{9} + \frac{Ma^2}{36} \]
Finding a common denominator (36):
\[ I_{\text{total}} = \frac{8Ma^2}{36} + \frac{Ma^2}{36} = \frac{9Ma^2}{36} = \frac{Ma^2}{4} \]

### Final Answer
The moment of inertia of the hollow cubical box about the specified axis is:
\[ \boxed{\frac{Ma^2}{4}} \]