Rotational Kinematics & Dynamics question from NEET 2004 (AIPMT)

Answer: (a)

1. Energy from lowest point to horizontal position (rise $l$): $v^2 = u^2 - 2gl$.
2. At the bottom $\vec u$ is horizontal; with the string horizontal $\vec v$ is vertical, so $\vec u \perp \vec v$.
3. $|\Delta \vec v| = \sqrt{u^2 + v^2} = \sqrt{u^2 + u^2 - 2gl} = \sqrt{2(u^2 - gl)}$
4. So the answer is (a) $\sqrt{2(u^2 - gl)}$.