Two bodies A and B have their moments of inertia $I_1$ and $I_2$ respectively about their axis of rotation.…

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
  1. $\frac{1}{3} \mathrm{I}_1$
  2. $\sqrt{3} \mathrm{I}_1$
  3. $\quad 2 \mathrm{I}_1$
  4. $3 \mathrm{I}_1$

Solution

$\begin{array}{ll} & (\text { K.E. })_A=(\text { K.E. })_B \\ & \frac{1}{2} \mathrm{I}_1 \omega_1^2=\frac{1}{2} I_2 \omega_2^2 \\ \therefore \quad & \frac{\omega_2^2}{\omega_1^2}=\frac{I_1}{I_2} ...(i)\\ & \text { Also, K.E }=\frac{1}{2} \mathrm{~L} \omega \\ \therefore \quad & \frac{1}{2} L_1 \omega_1=\frac{1}{2} L_2 \omega_2 \\ \therefore \quad & \frac{L_1}{L_2}=\frac{\omega_2}{\omega_1}=\frac{1}{\sqrt{3}} ...(ii)\\ \therefore \quad & \frac{I_1}{I_2}=\frac{1}{3} \\ \therefore \quad & I_2=3 I_1 \end{array}$ ....[From (i) and (ii)]

Asked in: MHT CET 2024 (10 May Shift 2)

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