From a disc of mass ' $M$ ' and radius ' $R$ ', a circular hole of diameter ' $R$ ' is cut whose rim passes…
- $\frac{13 \mathrm{MR}^2}{32}$
- $\frac{11 \mathrm{MR}^2}{32}$
- $\frac{9 \mathrm{MR}^2}{32}$
- $\frac{7 \mathrm{MR}^2}{32}$
Solution
Moment of inertia of disc is given by
$\mathrm{I}_{\text {disc }}=\mathrm{I}_{\mathrm{f}}+\mathrm{I}_{\text {hole }}$
.... $\left\{\mathrm{I}_{\mathrm{f}}=\right.$ M.I. of remaining part $\}$
$\begin{aligned}
& \therefore \quad \mathrm{I}_{\mathrm{r}}=\mathrm{I}_{\text {dise }}-\mathrm{I}_{\text {hole }} \\
& \mathrm{I}_{\text {disc }}=\frac{\mathrm{MR}^2}{2}
\end{aligned}$
By parallel axes theorem we get,
$\mathrm{I}_{\text {hole }}=\left[\frac{\frac{\mathrm{M}}{4}\left(\frac{\mathrm{R}}{2}\right)^2}{2}+\frac{\mathrm{M}}{4}\left(\frac{\mathrm{R}}{2}\right)^2\right]$
$\ldots\left\{\begin{array}{l}\because \mathrm{M}_{\text {bole }}=\frac{\mathrm{M}_{\text {dive }}}{4} \\ \because \text { the surface density is same }\end{array}\right\}$
$\therefore \quad \mathrm{I}_{\text {hole }}=\left[\frac{\mathrm{MR}^2}{32}+\frac{\mathrm{MR}^2}{16}\right]$
Substituting eq (iii) and eq (ii) in eq (i) we get,
$\begin{aligned}
\mathrm{I}_{\mathrm{r}} & =\frac{\mathrm{MR}^2}{2}-\frac{\mathrm{MR}^2}{32}-\frac{\mathrm{MR}^2}{16} \\
& =\mathrm{MR}^2\left[\frac{1}{2}-\frac{1}{32}-\frac{1}{16}\right] \\
& =\frac{13}{32} \mathrm{MR}^2
\end{aligned}$Asked in: MHT CET 2023 (11 May Shift 2)