Angular Momentum & its Conservation question from JEE Main 2019 (12 Apr, Shift 1)

Answer: (a)

1. At the lowest point the push is along the rope, so angular momentum about the pivot is conserved: $v' = \frac{vL}{L-\ell}$.
2. $\Delta K = \frac{1}{2}mv^2\left[\left(\frac{L}{L-\ell}\right)^2 - 1\right] \approx \frac{1}{2}mv^2\cdot\frac{2\ell}{L} = \frac{mv^2\ell}{L}$
3. $v^2 = 2gL(1-\cos\theta_0) \approx gL\theta_0^2$, so $\Delta K \approx mg\ell\theta_0^2$.
4. Work done $= mg\ell + \Delta K = mg\ell(1+\theta_0^2)$
5. So the answer is (a) $mg\ell(1+\theta_0^2)$.