The moment of inertia of a cube of mass m and side a about one of its edge is equal to ?

 

 

1. (8/3)ma2

2. (2/3)ma2

3. (4/3)ma2

4. (3)ma2

 

Dear student

We know that the moment of inertia of a square lamina about the axis passing through the centre of lamina and perpendicular to the plane of lamina is given by $I=\frac{M\left(a^2+a^2\right)}{12}=\frac{M{a}^2}{6}$.
Now assume that the thickness of the lamina is a so it becomes a cube but moment of inertia is not change so now the moment of inertia of the cube about the axis passing through an edge of the cube is given by the parallel axis theorem so
$$\begin{gathered} I_{edge}=\frac{M{a}^2}{6}+M{\left(\frac{a}{\sqrt{2}}\right)}^2 where \frac{a}{\sqrt{2}} is the dis\tan ce from the edge to the centre axis of cube \\ {I}_{edge}=\frac{M{a}^2}{6}+\frac{M{a}^2}{2} \\ {I}_{edge}=\frac{2M{a}^2}{3} \end{gathered}$$
Hence option (2) is correct.