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MHT CET · Physics · Rotational Motion

The moment of inertia of a circular disc of radius \(2 \mathrm{~m}\) and mass \(1 \mathrm{~kg}\) about an axis \(\mathrm{XY}\) passing through its center of mass and perpendicular to the plane of the disc is \(2 \mathrm{~kg} \mathrm{~m}^2\). The moment of inertia about an axis parallel to the axis \(\mathrm{XY}\) and passing through the edge of the disc is

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

Answer & Solution

Correct Answer

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

Step-by-step Solution

Detailed explanation

About XY axis, \(\mathrm{I}=\frac{1}{2} \mathrm{MR}^2=\frac{1}{2} \times 1 \times(2)^2=2 \mathrm{~kg} \cdot \mathrm{m}^2\)
Moment of inertia about an axis parallel to the axis XY and passing through the edge of the disc,
\(\mathrm{I}^{\prime}=\frac{1}{2} \mathrm{MR}^2+\mathrm{MR}^2=\frac{3}{2} \mathrm{MR}^2=6 \mathrm{~kg} \cdot \mathrm{m}^2\)