JEE Main · 2007 · 29 Apr 2007 · Offline · Conservation of Momentum

Take the centre of the larger disc of radius $2R$ as origin. The small disc of radius $R$ that is removed has one quarter of the area, hence mass $\frac{M}{4}$ if the full disc has mass $M$, and its centre lies at distance $R$ from the origin so that the circumferences just touch. Treating the hole as a negative mass, the centre of mass of what is left sits at $\frac{M\cdot 0-\frac{M}{4}\cdot R}{M-\frac{M}{4}}=-\frac{R}{3}$, a distance $\frac{R}{3}$ from the centre of the big disc, so $\alpha =\frac{1}{3}$.