MHT CET · Physics · Rotational Motion
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
- A \(\frac{\mathrm{MR}^{2}}{10}\)
- B \(\frac{\mathrm{MR}^{2}}{20}\)
- C \(\frac{2}{3} \mathrm{MR}^{2}\)
- D \(\frac{2}{5} \mathrm{MR}^{2}\)
Answer & Solution
Correct Answer
(B) \(\frac{\mathrm{MR}^{2}}{20}\)
Step-by-step Solution
Detailed explanation
\(\mathrm{I}_{1}=\frac{2}{5} \mathrm{MR}^{2}\)
\(\mathrm{I}_{2}=\frac{2}{5} \frac{M}{2} \times\left(\frac{R}{2}\right)^{2}=\frac{2}{5} \times \frac{M}{2} \times \frac{R^{2}}{4}=\frac{M R^{2}}{20}\)
\(\mathrm{I}_{2}=\frac{2}{5} \frac{M}{2} \times\left(\frac{R}{2}\right)^{2}=\frac{2}{5} \times \frac{M}{2} \times \frac{R^{2}}{4}=\frac{M R^{2}}{20}\)
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