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).
(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) | \(I_x=I_y\) |
1. | (a, c) | 2. | (a, d) |
3. | (b, d) | 4. | (b, c) |
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.
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