MHT CET · Physics · Rotational Motion
Two bodies A and B have their moments of inertia \(I_1\) and \(I_2\) respectively about their axis of rotation. If their kinetic energies of rotation are equal and their angular momenta \(\mathrm{L}_1\) and \(\mathrm{L}_2\) respectively are in the ratio \(1: \sqrt{3}\), then \(\mathrm{I}_2\) will be
- A \(\frac{1}{3} \mathrm{I}_1\)
- B \(\sqrt{3} \mathrm{I}_1\)
- C \(2 \mathrm{I}_1\)
- D \(3 \mathrm{I}_1\)
Answer & Solution
Correct Answer
(D) \(3 \mathrm{I}_1\)
Step-by-step Solution
Detailed explanation
\(\begin{array}{ll}
& (\text { K.E. })_A=(\text { K.E. })_B \\
& \frac{1}{2} \mathrm{I}_1 \omega_1^2=\frac{1}{2} I_2 \omega_2^2 \\
\therefore \quad & \frac{\omega_2^2}{\omega_1^2}=\frac{I_1}{I_2} ...(i)\\
& \text { Also, K.E }=\frac{1}{2} \mathrm{~L} \omega \\
\therefore \quad & \frac{1}{2} L_1 \omega_1=\frac{1}{2} L_2 \omega_2 \\
\therefore \quad & \frac{L_1}{L_2}=\frac{\omega_2}{\omega_1}=\frac{1}{\sqrt{3}} ...(ii)\\
\therefore \quad & \frac{I_1}{I_2}=\frac{1}{3} \\
\therefore \quad & I_2=3 I_1
\end{array}\)
....[From (i) and (ii)]
& (\text { K.E. })_A=(\text { K.E. })_B \\
& \frac{1}{2} \mathrm{I}_1 \omega_1^2=\frac{1}{2} I_2 \omega_2^2 \\
\therefore \quad & \frac{\omega_2^2}{\omega_1^2}=\frac{I_1}{I_2} ...(i)\\
& \text { Also, K.E }=\frac{1}{2} \mathrm{~L} \omega \\
\therefore \quad & \frac{1}{2} L_1 \omega_1=\frac{1}{2} L_2 \omega_2 \\
\therefore \quad & \frac{L_1}{L_2}=\frac{\omega_2}{\omega_1}=\frac{1}{\sqrt{3}} ...(ii)\\
\therefore \quad & \frac{I_1}{I_2}=\frac{1}{3} \\
\therefore \quad & I_2=3 I_1
\end{array}\)
....[From (i) and (ii)]
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