1. Given data

• Base of block on plane: $10\,\text{cm} \times 10\,\text{cm}$
• Height of block: $15\,\text{cm}$
• Coefficient of friction: $\mu = \sqrt{3}$
• Inclination $\theta$ is increased gradually from $0^\circ$

We must determine whether the block will slide first or topple first.

2. Condition for sliding

For a block on an incline, sliding begins when $mg\sin\theta = \mu mg\cos\theta$ so $\tan\theta = \mu = \sqrt{3}$ Hence, $\theta_s = 60^\circ$

So if sliding were to occur, it would start at $60^\circ$.

3. Condition for toppling

The block will topple when the vertical line through its center of mass passes through the lower edge of the base.

Take the cross-section in the plane of greatest slope.

• Base length along incline = $b = 10\,\text{cm}$
• Height perpendicular to incline = $h = 15\,\text{cm}$

The center of mass is at the center of the rectangle, i.e.

• $b/2 = 5\,\text{cm}$ from the lower edge along the plane
• $h/2 = 7.5\,\text{cm}$ above the plane

At toppling point, the vertical through the center of mass passes through the downhill edge, so $\frac{b}{2} = \frac{h}{2}\tan\theta$ Thus, $\tan\theta = \frac{b}{h} = \frac{10}{15} = \frac{2}{3}$ So, $\theta_t = \tan^{-1}\left(\frac{2}{3}\right) \approx 33.7^\circ$

4. Compare sliding angle and toppling angle

• Sliding angle: $\theta_s = 60^\circ$
• Toppling angle: $\theta_t \approx 33.7^\circ$

Since $\theta_t < \theta_s$

the block will topple before it gets a chance to slide.

Also, since friction is large enough, the block remains at rest until toppling occurs.

5. Check options

• A: At $60^\circ$, block starts sliding down plane. ❌ Incorrect, because it topples earlier at about $33.7^\circ$.
• B: Block remains at rest up to certain $\theta$ and then it topples. ✅ Correct.
• C: At $60^\circ$, it starts sliding and continues. ❌ Incorrect.
• D: At $60^\circ$, it starts sliding and then topples later. ❌ Incorrect.

6. Final answer

The correct option is: $\boxed{\text{B}}$

This matches the stored correct answer.