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 answer
    C
    $\frac{3}{2} M R^{2}$
  • 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'.

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