JEE Main · 2021 · 24 Feb 2021 · Shift 2 · Rotational Mechanics

Bending the bar into a regular hexagon gives six sides, each of length $2.4/6=0.4$ m and mass $6/6=1$ kg. Treat one side as a thin rod; its moment of inertia about an axis through its own centre and perpendicular to the plane of the hexagon is $\frac{1}{12}m{\ell }^2=\frac{1}{12}(1)(0.4{)}^2=\frac{0.16}{12}$ kg m${}^2$. The perpendicular distance from the centre of the hexagon to the midpoint of a side is the apothem $\frac{\sqrt{3}}{2}\ell =\frac{\sqrt{3}}{2}(0.4)$ m, whose square is $0.12$ m${}^2$, so by the parallel axis theorem each side contributes $\frac{0.16}{12}+1(0.12)=0.133$ kg m${}^2$ about the centre. The six identical sides add to $6\times 0.133=0.8$ kg m${}^2$, that is $8\times 10^{-1}$ kg m${}^2$.