Q130 / 160 · AP EAMCET 2008 · Physics

The moment of inertia of a thin circular disc about an axis passing through its centre and perpendicular to its plane is $I$. Then, the moment of inertia of the disc about an axis parallel to its diameter and touching the edge of the rim is

  • A
    $I$
  • B
    $2 I$
  • C
    $\frac{3}{2} I$
  • Correct answer
    D
    $\frac{5}{2} I$

Solution

Moment of inertia of a circular disc about an axis passing through centre of gravity and perpendicular to its plane $$ I=\frac{1}{2} M R^2 $$ From Eq. (i) $\quad M R^2=2 I$ Then, moment of inertia of disc about tangent $$ \begin{aligned} \text { in a plane }=\frac{5}{4} & M R^2 \\ & =\frac{5}{4}(2 I) \\ & =\frac{5}{2} I \end{aligned} $$

Hence, the correct answer is option: 'D'.

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