Q 11-06-104JEE MainJEE Main 2023 (30 Jan, Shift 1)Easy
A thin uniform rod of length $2\ \text{m}$, cross sectional area $A$ and density $d$ is rotated about an axis passing through the centre and perpendicular to its length with angular velocity $\omega$. If value of $\omega$ in terms of its rotational kinetic energy $E$ is $\sqrt{\dfrac{\alpha E}{Ad}}$, then the value of $\alpha$ is ______.
Numerical value type. Enter your answer.
Answer: 3
Mass $m=2Ad$, $I=\dfrac{mL^2}{12}=\dfrac{2Ad\times4}{12}=\dfrac{2Ad}{3}$.
$$E=\frac12I\omega^2=\frac{Ad}{3}\omega^2\ \Rightarrow\ \omega=\sqrt{\frac{3E}{Ad}}$$
So $\alpha=3$.
Solution by Sreeraj P, M.Sc Physics