JEE Main · 2009 · 12 Apr 2009 · Offline · Work, Power & Energy

At the lowest point the rod has its greatest angular speed $\omega$, and as it swings up this rotational energy is converted entirely into gravitational potential energy of the centre of mass, so $\frac{1}{2}I{\omega }^2=mgh$. For a rod pivoted at its end $I=\frac{1}{3}m{\ell }^2$, and substituting gives $\frac{1}{2}\cdot \frac{1}{3}m{\ell }^2{\omega }^2=mgh$, so the centre of mass rises to $h=\frac{{\ell }^2{\omega }^2}{6g}$.