A football of radius R is kept on a hole of radius r (𝑟 < 𝑅) made on a plank kept horizontally. One end of the plank is now lifted so that it gets tilted making an angle 𝜃 from the horizontal as shown in the figure below. The maximum value of $$\theta$$ so that the football does not start rolling down the plank satisfies (figure is schematic and not drawn to scale)

Consider a vertical plane that contains the line of steepest descent of the tilted plank. In this plane the problem reduces to a two-dimensional geometry of a circle (the football, radius $$R$$) resting on the two diametrically opposite edges of the hole (radius $$r$$).

Before tilting, the centre $$O$$ of the ball is a distance $$h = \sqrt{R^{2}-r^{2}}$$ below the rim of the hole.
After the plank is tilted through an angle $$\theta$$ about a horizontal axis (perpendicular to the plane we are considering), the two opposite edges of the hole acquire the coordinates

up-hill edge $$E_1 \, (-\,r\cos\theta,\; +\,r\sin\theta)$$
down-hill edge $$E_2 \, (+\,r\cos\theta,\; -\,r\sin\theta)$$

Let the instantaneous coordinates of the sphere’s centre be $$O\,(x,\;z)$$ in the same plane (origin at the hole’s centre, $$z$$ measured vertically upward).
Because the sphere is in contact with both edges, the centre must satisfy

$$\bigl(x + r\cos\theta\bigr)^{2} + \bigl(z - r\sin\theta\bigr)^{2} = R^{2} \;\;\;-(1)$$
$$\bigl(x - r\cos\theta\bigr)^{2} + \bigl(z + r\sin\theta\bigr)^{2} = R^{2} \;\;\;-(2)$$

Subtracting $$(2)$$ from $$(1)$$ gives

$$x\cos\theta - z\sin\theta = 0 \;\;\Longrightarrow\;\; x = z\tan\theta \;\;\;-(3)$$

Adding $$(1)$$ and $$(2)$$ yields

$$x^{2} + z^{2} + r^{2} = R^{2} \;\;\;-(4)$$

Using $$(3)$$ in $$(4)$$:

$$z^{2}\tan^{2}\theta + z^{2} + r^{2} = R^{2}$$
$$\frac{z^{2}}{\cos^{2}\theta} + r^{2} = R^{2}$$
$$z^{2} = \bigl(R^{2}-r^{2}\bigr)\cos^{2}\theta \;\;\;-(5)$$

Therefore

$$x^{2} = z^{2}\tan^{2}\theta = \bigl(R^{2}-r^{2}\bigr)\sin^{2}\theta \;\;\;-(6)$$

The ball will be on the verge of rolling out when the vertical line through its centre just passes through the down-hill edge $$E_2$$. At this limiting case

$$x = r\cos\theta \;\;\;-(7)$$

Substituting $$(6)$$ into $$(7)$$:

$$\sqrt{R^{2}-r^{2}}\;\sin\theta = r\cos\theta$$
$$\tan\theta = \frac{r}{\sqrt{R^{2}-r^{2}}} \;\;\;-(8)$$

Using $$\sin\theta = \dfrac{\tan\theta}{\sqrt{1+\tan^{2}\theta}}$$ and simplifying with $$(8)$$:

$$\sin\theta = \frac{r}{\sqrt{R^{2}-r^{2}}}\;\cdot\; \frac{\sqrt{R^{2}-r^{2}}}{R} = \frac{r}{R}$$

Hence the maximum tilt angle is determined by

$$\boxed{\sin\theta = \dfrac{r}{R}}$$

Option A which is: $$\sin \theta = \dfrac{r}{R}$$