The moment of inertia of a circular disc of radius $2 \mathrm{~m}$ and mass $1 \mathrm{~kg}$ about an axis passing through the centre of mass but perpendicular to the plane of the disc is $2 \mathrm{~kg} \mathrm{~m}^{2}$. Its moment of inertia about an axis parallel to this axis but passing through the edge of the disc is (see the given figure).

• A

  $8 \mathrm{~kg} \mathrm{~m}^{2}$
• B

  $4 \mathrm{~kg} \mathrm{~m}^{2}$
• C

  $10 \mathrm{~kg} \mathrm{~m}^{2}$
• Correct answer
  D

  $6 \mathrm{~kg} \mathrm{~m}^{2}$

Solution

According to parallel axis theorem $$ \begin{aligned} \mathrm{I}_{X^{\prime} Y^{\prime}} &=I_{X Y}+M^{2} \\ &=2+(1)(2)^{2} \\ &=6 \mathrm{~kg}-\mathrm{m}^{2} \end{aligned} $$

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