Physics - Rotation Question with Solution | TestHub

${\mathrm{I}}_{\mathrm{remaining}}={\mathrm{I}}_{\mathrm{total}}-{\mathrm{I}}_{\mathrm{removed}}$

${\mathrm{I}}_{\mathrm{remaining}}={\mathrm{I}}_{\mathrm{total}}-4{\mathrm{I}}_{\mathrm{each} \mathrm{disc}}$
${\mathrm{I}}_{\mathrm{remaining}}=\mathrm{I}={\mathrm{I}}_{\mathrm{total}}-4\times ({\mathrm{I}}_{\mathrm{disc}}+{\mathrm{M}}_{\mathrm{disc}}{\left(\sqrt{2}\mathrm{R}\right)}^2)$  (From parallel axis theorem)
$\mathrm{I}=\frac{1}{12}\mathrm{M}\left(32{\mathrm{R}}^2\right)-4\times \left(\frac{1}{2}\times \frac{\mathrm{\pi M}}{16}\times {\mathrm{R}}^2+\frac{\mathrm{\pi M}}{16}\times 2{\mathrm{R}}^2\right)$               $\left(\because {\mathrm{M}}_{\mathrm{disc}}=\frac{\mathrm{\pi M}}{16}\right)$
$\mathrm{I}=\frac{8}{3}{\mathrm{MR}}^2-4\times \frac{5}{2}\times \frac{\mathrm{\pi M}}{16}\times {\mathrm{R}}^2={\mathrm{MR}}^2\left(\frac{8}{3}-\frac{20\mathrm{\pi }}{32}\right)$
$\mathrm{I}=$$\left(\frac{8}{3}-\frac{10\mathrm{\pi }}{16}\right)\mathrm{M}{\mathrm{R}}^2$