The moment of inertia of a uniform circular disc of radius ' $R$ ' and mass ' $M$ ' about an axis passing…
- $\mathrm{MR}^{2}$
- $\frac{1}{2} \mathrm{MR}^{2}$
- $\frac{3}{2} \mathrm{MR}^{2}$
- $\frac{7}{2} \mathrm{MR}^{2}$
Solution
$\mathrm{I}_{\mathrm{C} . \mathrm{M}}=\frac{1}{2} \mathrm{MR}^{2}$

From parallel axis theorem $\mathrm{I}_{\mathrm{T}}=\mathrm{I}_{\mathrm{C} . \mathrm{M} .}+\mathrm{MR}^{2}=\frac{1}{2} \mathrm{MR}^{2}+\mathrm{MR}^{2}=\frac{3}{2} \mathrm{MR}^{2}$
Asked in: JEE Mains - Rotational Motion - Test 1