A uniform thin rod of 120 cm length and 1600 g mass is bent as shown in the figure. The moment of inertia of the bent rod about an axis passing through the point 'O' and perpendicular to the plane of the paper is ________ kg m².

There are four rods, each weighing 400 g$=\frac{400}{1000}=0.4\mathrm{kg}$

length l = $30\mathrm{cm}=\frac{30}{100}\mathrm{m}=0.3\mathrm{m}$

Now that both sets of two rods have the same M.I, we can compute the moment of inertia using the following formula.

${\mathrm{h}}^2={\mathrm{l}}^2-{\left(\frac{\mathrm{l}}{2}\right)}^2=\frac{3}{4}{\mathrm{l}}^2$

moment of inertia of the system,

$\mathrm{M}.\mathrm{I} \mathrm{system} =( \mathrm{MI} \mathrm{of} \mathrm{I})\times 2+(\mathrm{M}. \mathrm{I} \mathrm{of} \mathrm{II})\times 2$

$=\frac{\mathrm{m}{l}^2}{3}\times 2+2\left(\frac{\mathrm{m}{l}^2}{12}+{\mathrm{mh}}^2\right)$

$=\frac{2}{3}\mathrm{m}{l}^2+2\left(\frac{\mathrm{m}{l}^2}{12}+\frac{3}{4}\mathrm{m}{l}^2\right)$

$=\frac{2}{3}\mathrm{m}{l}^2+2\mathrm{m}{l}^2\frac{10}{12}$

$=\frac{7}{3}\mathrm{m}{l}^2=\frac{7}{3}\times 0.4\times 0.3\times 0.3=0.084{\mathrm{kgm}}^2$

Hence the correct answer is 0.084.