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.

Answer

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 Geometry of the Cube
The hollow cubical box has six square faces. The axis of rotation is passing through the centers of two opposite faces.

### Step 2: Moment of Inertia of One Face
The moment of inertia \( I \) of a square plate about an axis passing through its center and perpendicular to its plane is given by:
\[ I_{\text{face}} = \frac{1}{12} m a^2 \]
where \( m \) is the mass of the square face and \( a \) is the side length of the square.

### Step 3: Mass of Each Face
Since the total mass \( M \) is distributed equally among the six faces, the mass of each face is:
\[ m = \frac{M}{6} \]

### Step 4: Calculate Moment of Inertia for One Face
Substituting \( m \) into the moment of inertia formula for one face, we get:
\[ I_{\text{face}} = \frac{1}{12} \left(\frac{M}{6}\right) a^2 = \frac{M a^2}{72} \]

### Step 5: Total Moment of Inertia for Two Faces
Since there are two faces contributing to the moment of inertia about the axis, we multiply by 2:
\[ I_1 = 2 \times I_{\text{face}} = 2 \times \frac{M a^2}{72} = \frac{M a^2}{36} \]

### Step 6: Moment of Inertia of the Side Faces
For the four side faces, we need to use the parallel axis theorem. The moment of inertia of a face about an axis through its center is:
\[ I_{\text{side}} = I_{\text{cm}} + Md^2 \]
where \( I_{\text{cm}} = \frac{1}{12} m a^2 \) and \( d = \frac{a}{2} \) (the distance from the center of mass to the axis).

### Step 7: Calculate Moment of Inertia for One Side Face
Using the mass of each side face \( m = \frac{M}{6} \):
\[ I_{\text{side}} = \frac{1}{12} \left(\frac{M}{6}\right) a^2 + \left(\frac{M}{6}\right) \left(\frac{a}{2}\right)^2 \]
Calculating this gives:
\[ I_{\text{side}} = \frac{M a^2}{72} + \frac{M a^2}{24} = \frac{M a^2}{72} + \frac{3M a^2}{72} = \frac{4M a^2}{72} = \frac{M a^2}{18} \]

### Step 8: Total Moment of Inertia for Four Side Faces
Now, for the four side faces:
\[ I_2 = 4 \times I_{\text{side}} = 4 \times \frac{M a^2}{18} = \frac{4M a^2}{18} = \frac{2M a^2}{9} \]

### Step 9: Total Moment of Inertia
Finally, the total moment of inertia \( I \) about the axis is:
\[ I = I_1 + I_2 = \frac{M a^2}{36} + \frac{2M a^2}{9} \]
Finding a common denominator (36):
\[ I = \frac{M a^2}{36} + \frac{8M a^2}{36} = \frac{9M a^2}{36} = \frac{M a^2}{4} \]

### Final Result
Thus, the moment of inertia of the hollow cubical box about the specified axis is:
\[ \boxed{\frac{M a^2}{4}} \]