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Question: Answered & Verified by Expert
If the spherical planet of mass ' $\mathrm{M}^{\prime}$ and radius 'R' suddenly shrinks to half its size, its mass reduces to half. The new moment of inertia of the planet about its diameter is
PhysicsRotational MotionMHT CETMHT CET 2020 (20 Oct Shift 2)
Options:
  • A $\frac{\mathrm{MR}^{2}}{10}$
  • B $\frac{\mathrm{MR}^{2}}{20}$
  • C $\frac{2}{3} \mathrm{MR}^{2}$
  • D $\frac{2}{5} \mathrm{MR}^{2}$
Solution:
2544 Upvotes Verified Answer
The correct answer is: $\frac{\mathrm{MR}^{2}}{20}$
$\mathrm{I}_{1}=\frac{2}{5} \mathrm{MR}^{2}$
$\mathrm{I}_{2}=\frac{2}{5} \frac{M}{2} \times\left(\frac{R}{2}\right)^{2}=\frac{2}{5} \times \frac{M}{2} \times \frac{R^{2}}{4}=\frac{M R^{2}}{20}$

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