Q105 / 150 · MHT CET 2007 · Physics
The moment of inertia of a uniform circular disc of radius $R$ and mass $M$ about an axis touching the disc at its diameter and normal to the disc is
- A$M R^{2}$
- B$\frac{2}{5} M R^{2}$
- Correct answerC$\frac{3}{2} M R^{2}$
- D$\frac{1}{2} M R^{2}$
Solution
The moment of inertia about an axis passing through centre of mass of disc and perpendicular to its plane is
$$
I_{\mathrm{CM}}=\frac{1}{2} M R^{2}
$$
Where $M$ is the mass of disc and $R$ its radius According to theorem of parallel axis, moment of inertia of circular disc about an axis
touching the disc at its diameter and normal to the disc is
$$
\begin{aligned}
I &=I_{\mathrm{CM}}+M R^{2} \\
&=\frac{1}{2} M R^{2}+M R^{2} \\
&=\frac{3}{2} M R^{2}
\end{aligned}
$$
Hence, the correct answer is option: 'C'.