A ring is hung on a nail. It can oscillate, without slipping or sliding (i) in its plane with a time period $$T_1$$ and (ii) back and forth in a direction perpendicular to its plane, with a period $$T_2$$. The ratio $$\frac{T_1}{T_2}$$ will be:

Let the ring have mass $$M$$ and radius $$R$$. The nail goes through a point on the circumference, so the distance from the pivot to the centre of mass is simply $$d = R$$.

For any rigid body oscillating with small amplitude about a fixed horizontal axis, the time period is given by the physical-pendulum formula

$$T = 2\pi \sqrt{\dfrac{I_P}{M g d}},$$

where $$I_P$$ is the moment of inertia about the chosen pivot axis and $$d$$ is the distance between the pivot and the centre of mass.

We now evaluate $$I_P$$ for the two different oscillations.

Oscillation in the plane of the ring : time period $$T_1$$

If the ring swings in its own plane, the axis through the nail is perpendicular to the plane. For a thin ring, the moment of inertia about the axis through the centre and perpendicular to the plane is

$$I_C = M R^{2}.$$

Using the parallel-axis theorem, $$I_{P1} = I_C + M R^{2} = M R^{2} + M R^{2} = 2 M R^{2}.$$

Substituting in the time-period formula, we obtain

$$T_1 = 2\pi \sqrt{\dfrac{2 M R^{2}}{M g R}} = 2\pi \sqrt{\dfrac{2 R}{g}}.$$

Oscillation perpendicular to the plane of the ring : time period $$T_2$$

Here the ring tilts forward and backward, so the axis through the nail lies in the plane of the ring and is tangential to the circle. First, for a thin ring the moment of inertia about any diameter (an axis in the plane through the centre) is, by the perpendicular-axis theorem,

$$I_\text{diameter} = \dfrac{1}{2} M R^{2}.$$

The tangent axis is parallel to this diameter but displaced by a distance $$R$$, so using the parallel-axis theorem again,

$$I_{P2} = I_\text{diameter} + M R^{2} = \dfrac{1}{2} M R^{2} + M R^{2} = \dfrac{3}{2} M R^{2}.$$

Therefore

$$T_2 = 2\pi \sqrt{\dfrac{\dfrac{3}{2} M R^{2}}{M g R}} = 2\pi \sqrt{\dfrac{3 R}{2 g}}.$$

Ratio of the two periods

Dividing the two expressions, the common factors $$2\pi$$, $$M$$ and $$R/g$$ cancel out:

$$\dfrac{T_1}{T_2} \;=\; \sqrt{\dfrac{I_{P1}}{I_{P2}}} \;=\; \sqrt{\dfrac{2 M R^{2}}{\dfrac{3}{2} M R^{2}}} = \sqrt{\dfrac{2}{\dfrac{3}{2}}} = \sqrt{\dfrac{4}{3}} = \dfrac{2}{\sqrt{3}}.$$

Hence, the correct answer is Option A.