With reference to the figure of a cube of edge \(a\) and mass \(m,\) the following statements are given. (\(O\) is the centre of the cube).

    

Choose the correct statements:
(a) The moment of inertia of the cube about the \(z\)-axis is \(I_z=I_x+I_y\)
(b)  The moment of inertia of the cube about the \(z'\)-axis is \({I_z}'=I_z+\frac{{ma}^2}{2}\)
(c) The moment of inertia of the cube about \(z''\)-axis is = \(=I_z+\frac{{ma}^2}{2}\)
(d) \(I_x=I_y\)
Choose the correct option from the given options:
1. (a, c) 2. (a, d)
3. (b, d) 4. (b, c)
 

Hint: Apply the parallel axis theorem  \({I}={I}_{{cm}}+{mr}^2\)

Explanation: Let us analyse each statement one by one.

In case (a); 
The theorem of perpendicular axes applies only to laminar (plane) objects, and in the given case, it is applied along a 3D cube. Thus, this is false.
In case (b); 
\(z'\) is a parallel axis at a distance of \(\frac{a}{\sqrt 2}\) from \(z.\) By the parallel axis theorem,
\(\Rightarrow I_{z'}=I_z+m\left(\frac{a}{\sqrt{2}}\right)^2=I_z+\frac{m a^2}{2}\)
Thus, the given statement is true.
In case (c);
\(z"\) is not parallel to \(z.\) Thus, the theorem of the parallel axis cannot be applied. Thus, this is false.
In case (d);
As \(x\) and \(y\)-axes are symmetrical, the inertia about them will be equal, \(I_x=I_y.\) Thus, this statement is true.
Therefore, statements (b) and (d) are true for the given conditions.
Hence, option (3) is the correct answer.