Q140 / 150 · MHT CET 2021 (21 Sep Shift 1) · Physics
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
- Correct answerA$6 \mathrm{~kg} \mathrm{~m}^2$
- B$4 \mathrm{~kg} \mathrm{~m}^2$
- C$10 \mathrm{~kg} \mathrm{~m}^2$
- D$8 \mathrm{~kg} \mathrm{~m}^2$
Solution
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,
$$
\mathrm{I}^{\prime}=\frac{1}{2} \mathrm{MR}^2+\mathrm{MR}^2=\frac{3}{2} \mathrm{MR}^2=6 \mathrm{~kg} \cdot \mathrm{m}^2
$$
Hence, the correct answer is option: 'A'.