A uniform thin bar of mass 6 kg and length 2.4 meter is bent to make an equilateral hexagon. The moment of inertia about an axis passing through the centre of mass and perpendicular to the plane of hexagon is ______ $$\times 10^{-1}$$ kg m$$^2$$.

A uniform thin bar of mass $$M = 6$$ kg and length $$L = 2.4$$ m is bent into a regular hexagon. Since a regular hexagon has 6 equal sides, the length of each side is $$a = \frac{L}{6} = \frac{2.4}{6} = 0.4$$ m, and the mass of each side is $$m = \frac{M}{6} = \frac{6}{6} = 1$$ kg.

For a regular hexagon with side $$a$$, the perpendicular distance from the centre to the midpoint of each side is $$d = \frac{\sqrt{3}}{2} a = \frac{\sqrt{3}}{2} \times 0.4 = 0.2\sqrt{3}$$ m.

The moment of inertia of each side (a thin rod of mass $$m$$ and length $$a$$) about the axis through the centre of the hexagon and perpendicular to its plane is found using the parallel axis theorem. Each rod has its own moment of inertia about its centre given by $$\frac{ma^2}{12}$$, and using the parallel axis theorem with distance $$d$$, the moment of inertia of each rod about the central axis is $$I_{\text{rod}} = \frac{ma^2}{12} + md^2$$.

Substituting the values: $$I_{\text{rod}} = \frac{1 \times (0.4)^2}{12} + 1 \times (0.2\sqrt{3})^2 = \frac{0.16}{12} + 0.12 = 0.01333 + 0.12 = 0.13333$$ kg m$$^2$$.

Since the hexagon has 6 identical sides, the total moment of inertia is $$I = 6 \times I_{\text{rod}} = 6 \times 0.13333 = 0.8$$ kg m$$^2$$.

Expressing this as $$x \times 10^{-1}$$ kg m$$^2$$: $$0.8 = 8 \times 10^{-1}$$.

Therefore, the answer is $$8$$.