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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).

  1. A \(8 \mathrm{~kg} \mathrm{~m}^{2}\)
  2. B \(4 \mathrm{~kg} \mathrm{~m}^{2}\)
  3. C \(10 \mathrm{~kg} \mathrm{~m}^{2}\)
  4. D \(6 \mathrm{~kg} \mathrm{~m}^{2}\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(6 \mathrm{~kg} \mathrm{~m}^{2}\)

Step-by-step Solution

Detailed explanation

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