MHT CET · Physics · Rotational Motion
Two rotating bodies \(\mathrm{P}\) and \(\mathrm{Q}\) of masses ' \(\mathrm{m}\) ' with moment of inertia \(I_P\) and \(I_Q\left(I_Q>I_P\right)\) have equal Kinetic energy of rotation. If \(L_P\) and \(L_Q\) be their angular momenta respectively then
- A \(\mathrm{L}_{\mathrm{Q}}=0\)
- B \(\mathrm{L}_{\mathrm{Q}}=\mathrm{L}_{\mathrm{P}}\)
- C \(\mathrm{L}_{\mathrm{Q}} < \mathrm{L}_{\mathrm{P}}\)
- D \(\mathrm{L}_{\mathrm{Q}}>\mathrm{L}_{\mathrm{P}}\)
Answer & Solution
Correct Answer
(D) \(\mathrm{L}_{\mathrm{Q}}>\mathrm{L}_{\mathrm{P}}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \text { Kinetic energy }=\frac{\mathrm{L}_{\mathrm{P}}^2}{2 \mathrm{I}_{\mathrm{P}}}=\frac{\mathrm{L}_{\mathrm{Q}}^2}{\mathrm{LI}_{\mathrm{Q}}} \\ & \therefore \frac{\mathrm{L}_{\mathrm{Q}}^2}{\mathrm{~L}_{\mathrm{P}}^2}=\frac{\mathrm{I}_{\mathrm{Q}}}{\mathrm{I}_{\mathrm{P}}} \\ & \because \mathrm{I}_{\mathrm{Q}}>\mathrm{I}_{\mathrm{P}} \text { we get } \mathrm{L}_{\mathrm{Q}}>\mathrm{L}_{\mathrm{P}}\end{aligned}\)
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