Physics - Rotation Question with Solution | TestHub

If $M$ is mass of the square plate before cutting the holes, then mass of portion of each hole, $m=\frac{M}{16R^2}\times \pi R^2=\frac{M\pi }{16}$ $\therefore$ Moment of inertia of remaining portion about $Z$ axis $$\begin{array}{rl}& I=I_{\text{square }}-4I_{\text{hole }}\\ & =\frac{M}{12}\left(16R^2+16R^2\right)-4\left[\frac{m{R}^2}{2}+m(\sqrt{2}R{)}^2\right]\\ & =\frac{M}{12}\times 32R^2-10m{R}^2\\ & =\frac{8}{3}M{R}^2-\frac{10}{16}M{R}^2\pi \\ & I=\left(\frac{8}{3}-\frac{10\pi }{16}\right)M{R}^2.\end{array}$$