A disc of mass ‘m’ and radius R has a concentric hole of radius ‘r’. Its M.I. about an axis through centre and normal to its plane is

# A disc of mass ‘m’ and radius R has a concentric hole of radius ‘r’. Its M.I. about an axis through centre and normal to its plane is

1. A

$\frac{\mathrm{m}}{2}{\left(\mathrm{R}-\mathrm{r}\right)}^{2}$

2. B

$\frac{\mathrm{m}}{2}\left({\mathrm{R}}^{2}-{\mathrm{r}}^{2}\right)$

3. C

$\frac{\mathrm{m}}{2}{\left(\mathrm{R}+\mathrm{r}\right)}^{2}$

4. D

$\frac{\mathrm{m}}{2}\left({\mathrm{R}}^{2}+{\mathrm{r}}^{2}\right)$

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### Solution:

The disc's moment of inertia when it has a hole

${\mathrm{I}}_{\mathrm{ο}}=\frac{1}{2}{\mathrm{m}}_{1}{\mathrm{R}}^{2}-\frac{1}{2}{\mathrm{m}}_{2}{\mathrm{r}}^{2}$

${\mathrm{m}}_{1}-{\mathrm{m}}_{2}=\mathrm{m______}\left(1\right)$

$\mathrm{also},$

$⇒{\mathrm{m}}_{1}\left(1-\frac{{\mathrm{r}}^{2}}{{\mathrm{R}}^{2}}\right)=\mathrm{m}$

${\mathrm{m}}_{1}=\frac{{\mathrm{mR}}^{2}}{{\mathrm{R}}^{2}-{\mathrm{r}}^{2}}$

${\mathrm{m}}_{2}=\frac{{\mathrm{mr}}^{2}}{{\mathrm{R}}^{2}-{\mathrm{r}}^{2}}$

${\mathrm{I}}_{\mathrm{ο}}=\frac{\mathrm{m}}{2\left({\mathrm{R}}^{2}-{\mathrm{r}}^{2}\right)}\left({\mathrm{R}}^{4}-{\mathrm{r}}^{4}\right)=\frac{\mathrm{m}}{2}\left({\mathrm{R}}^{2}+{\mathrm{r}}^{2}\right)$

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