1. Understand the geometry

A bigger disc has radius $2R$. A smaller disc of radius $R$ is removed.

Since the circumferences coincide internally, the smaller disc lies inside the bigger one and touches it internally. So the distance between their centers is:

$d = 2R - R = R$

Let the center of the bigger disc be at $O$ and the center of the removed smaller disc be at $C$. Then:

$OC = R$

2. Use center of mass of a system with negative mass

Assume uniform surface mass density $\sigma$.

• Mass of bigger disc: $M_1 = \sigma \pi (2R)^2 = 4\sigma\pi R^2$

• Mass of removed smaller disc: $M_2 = \sigma \pi R^2$

Treat the removed disc as negative mass located at distance $R$ from the center of the bigger disc.

Net mass:

$M = M_1 - M_2 = 4\sigma\pi R^2 - \sigma\pi R^2 = 3\sigma\pi R^2$

3. Find the center of mass of the remaining lamina

Take origin at the center of the bigger disc. The bigger disc's center of mass is at $x=0$. The removed disc has center at $x=R$.

So,

$x_{\text{CM}} = \frac{M_1(0) - M_2(R)}{M_1 - M_2}$

$x_{\text{CM}} = \frac{0 - (\sigma\pi R^2)(R)}{3\sigma\pi R^2} = -\frac{R}{3}$

The negative sign means the new center of mass lies on the side opposite to the removed part.

Hence its distance from the center of the bigger disc is:

$\frac{R}{3}$

So,

$\alpha = \frac{1}{3}$

4. Check options

• A: $\frac14$ ❌
• B: $\frac13$ ✅
• C: $\frac12$ ❌
• D: $\frac16$ ❌

Therefore, the correct option is:

$\boxed{\frac{1}{3}}$

So Option B is correct.