A thin square plate has mass M and side L. Four holes, each of radius R, are cut out as shown in Fig. The ratio of moment of inertia of the remaining portion to the moment of inertia of the complete plate is

 

 

Mass per unit are of the plate is $\sigma =\frac{M}{L^2}$

Mass of each hole is $m=\sigma \times {\mathrm{\pi R}}^2=\frac{{\mathrm{\pi MR}}^2}{{\mathrm{L}}^2}$

Since L = 4R, m = $\frac{\mathrm{\pi M}}{16}$

Distance of the centre of a hole and centre O of the plate is 

            x = AO = $\sqrt{2}$R 

Using parallel - axes theorem, the moment of inertia of a hole about the z-axis is 

                     $$\begin{gathered} I=\frac{1}{2}m{R}^2+m{x}^2 \\ =m\left(\frac{R^2}{2}+2R^2\right) \\ =\frac{\mathrm{\pi M}}{32}\times \frac{5R^2}{2}=\frac{5{\mathrm{\pi MR}}^2}{64} \end{gathered}$$

Moment of four holes about the z-axis=4I= $\frac{5{\mathrm{\pi MR}}^2}{8}$

Moment of inertia of the complete square plate is  $I_1=\frac{M{L}^2}{6}=\frac{M}{6}\times {\left(4R\right)}^2=\frac{8}{3}M{R}^2$.

Therefore, moment of inertia of the remaining portion about the z-axis is $I_2=\frac{8}{3}M{R}^2-\frac{5\mathrm{\pi }}{8}M{R}^2=\left(\frac{8}{3}-\frac{5\mathrm{\pi }}{8}\right)M{R}^2$

The required ratio =$\frac{I_2}{I_1}=1-\frac{15\mathrm{\pi }}{64}=0.26$