Consider a uniform rod of mass $M = 4m$ and length $l$ pivoted about its centre. A mass $m$ moving with velocity $v$ making angle $\theta = \dfrac\pi4$ to the rod's long axis collides with one end of the rod and sticks to it. The angular speed of the rod-mass system just after the collision is:
Answer: (C) $\dfrac{3\sqrt2}{7}\dfrac vl$
Angular momentum about the pivot is conserved. Only the velocity component perpendicular to the rod, $v\sin45^\circ$, contributes:
$$L = m\,v\sin45^\circ\cdot\frac l2 = \frac{mvl}{2\sqrt2}$$
$$I = \frac{(4m)l^2}{12} + m\left(\frac l2\right)^2 = \frac{ml^2}{3} + \frac{ml^2}{4} = \frac{7ml^2}{12}$$
$$\omega = \frac LI = \frac{mvl}{2\sqrt2}\cdot\frac{12}{7ml^2} = \frac{6}{7\sqrt2}\frac vl = \frac{3\sqrt2}{7}\frac vl$$
(The official answer key lists $\frac37\frac vl$, which drops the factor $\sqrt2$.)
Solution by Sreeraj P, M.Sc Physics