Rotational Kinematics & Dynamics question from JEE Main 2019 (12 Apr, Shift 1)

Answer: (b)

1. Tension at $x$ provides centripetal force for the part from $x$ to $\ell$: $T(x) = \int_x^{\ell}\dfrac{m}{\ell}\omega^2 s\,ds$.
2. $T(x) = \dfrac{m\omega^2}{2\ell}(\ell^2 - x^2)$: a downward parabola, maximum at $x = 0$ with zero slope, zero at $x = \ell$.
3. This matches graph (b).