M.I of lamina = (M.I of rectangle about base AB) - (M.I of semi-circle about base AB)

$$ \begin{array}{l}{=\left(\mathrm{I}_{\mathrm{G}}+\mathrm{Ah}^{2}\right)_{\mathrm{I}}-\left(\mathrm{I}_{\mathrm{G}}+\mathrm{Ah}^{2}\right)_{\mathrm{II}}} \\ {=\left(\frac{\mathrm{bd}^{3}}{12}+(\mathrm{bd}) \times\left(\frac{\mathrm{d}}{2}\right)^{2}\right)_{\mathrm{I}}-\left(0.11 \mathrm{R}^{4}+\left(\frac{\pi \mathrm{d}^{2}}{8}\right) \times\left(\frac{4 \mathrm{R}}{3 \pi}\right)^{2}\right)_{\mathrm{II}}}\end{array} $$

$$ \begin{array}{l}{=\left(\frac{120 \times 500^{3}}{12}+(120 \times 500) \times\left(\frac{500}{2}\right)^{2}\right)_{1}-\left(0.11 \times 60^{4}+\left(\frac{\pi \times 120^{2}}{8}\right) \times\left(\frac{4 \times 60}{3 \pi}\right)^{2}\right)_{\mathrm{II}}} \\ {=\left(5 \times 10^{9}\right)_{\mathrm{I}}+\left(5.09 \times 10^{6}\right)_{\mathrm{II}}} \\ {=4.99 \times 10^{9} \mathrm{mm}^{4}}\end{array} $$