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MHT CET · Physics · Rotational Motion

Two bodies rotate with kinetic energies ' \(E_1\) ' and ' \(E_2\) '. Moment of inertia about their axis of rotation are ' \(I_1\) ' and ' \(I_2\) '. If \(I_1=\frac{I_2}{3}\) and \(\mathrm{E}_1=27 \mathrm{E}_2\), then the ratio of angular momenta ' \(\mathrm{L}_1\) ' to ' \(\mathrm{L}_2\) ' is

  1. A \(1: 3\)
  2. B \(3: 1\)
  3. C \(1: 1\)
  4. D \(2: 1\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(3: 1\)

Step-by-step Solution

Detailed explanation

\( \begin{aligned} & \mathrm{E}_1=\frac{1}{2} \mathrm{I}_1 \omega_1^2 \\ & \mathrm{E}_2=\frac{1}{2} \mathrm{I}_2 \omega_2^2=\frac{1}{2}\left(3 \mathrm{I}_1\right) \omega_2^2=\frac{3}{2} \mathrm{I}_1 \omega_2^2 \quad\left[\because \mathrm{I}_2=3 \mathrm{I}_1\right] \\ & \mathrm{E}_1=27 \mathrm{E}_2 \\ & \frac{1}{2} \mathrm{I}_1 \omega_1^2=\frac{81}{2} \mathrm{I}_1 \omega_2^2 \\ & \omega_1=9 \omega_2 \\ & \frac{\mathrm{L}_1}{\mathrm{~L}_2}=\frac{\mathrm{I}_1 \omega_1}{\mathrm{I}_2 \omega_2}=\frac{\mathrm{I}_1 \times 9 \omega_2}{3 \mathrm{I}_1 \times \omega_2}=3 \end{aligned} \)
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