JEE Physics 2023 Question with Solution

Given: A thin uniform rod has length $2 \, \text{m}$, cross-sectional area $A$, density $d$, and rotates about an axis through its center perpendicular to its length with angular velocity $\omega$.

Find: The value of $\alpha$ in the relation

$$\omega = \frac{\alpha E}{Ad}$$

using the final answer indicated by the solution.

The solution is inconsistent with the question statement. It contains a completely unrelated ac generator expression,

$$\xi_{\max} = NAB\omega$$

and concludes with $1584$, which does not correspond to the rod-rotation question.

Since the page itself explicitly states Correct Answer: 1584, the extracted numerical answer is taken as $1584$.

Therefore, the value of $\alpha$ is $1584$.