KCET · Physics · Rotational Motion
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 \(10 \mathrm{~kg} \mathrm{~m}^{2}\)
- D \(6 \mathrm{~kg} \mathrm{~m}^{2}\)
Answer & Solution
Correct Answer
(D) \(6 \mathrm{~kg} \mathrm{~m}^{2}\)
Step-by-step Solution
Detailed explanation
According to parallel axis theorem
\(\begin{aligned}\mathrm{I}_{X^{\prime} Y^{\prime}} &=I_{X Y}+M^{2} \\&=2+(1)(2)^{2} \\&=6 \mathrm{~kg}-\mathrm{m}^{2}\end{aligned}\)

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

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