Q108 / 160 · JIPMER 2013 · Physics
Two rigid bodies $A$ and $B$ rotate with rotational kinetic energies $E_A$ and $E_B$ respectively. The moments of inertia of $A$ and $B$ about the axis of rotation are $I_A$ and $I_B$ respectively.
If $I_A=\frac{I_B}{4}$ and $E_A=100=E_B$, the ratio of angular momentum $\left(L_A\right)$ of $A$ to the angular momentum $\left(L_B\right)$ of $B$ is
- A25
- B$5 / 4$
- Correct answerC5
- D$1 / 4$
Solution
From, the formula, rotational KE
$E_{\text {rotational }}=\frac{1}{2} l \omega^2=\frac{L^2}{2 \mid}(\because L=\mid \omega)$
Therefore, $\quad L^2=2 E l$
Hence
$L_A^2=2 E_A I_A\ldots(i)$
and
$L_B^2=2 E_B I_B\ldots(ii)$
From Eq. (i) and (ii)
$\begin{aligned} & \frac{L_A}{L_B}=\sqrt{\frac{2 E_A I_B}{2 E_B I_A}}=\sqrt{\frac{100}{4}}=5 \\ & \quad\left(\because \frac{E_B}{E_A}=100 \text { and } \frac{I_A}{I_B}=4\right)\end{aligned}$
Hence, the correct answer is option: 'C'.