Toppling versus Sliding of a Block on an Increasing Incline

To determine whether the block slides or topples first as the angle of inclination $\theta$ is gradually increased, we evaluate the critical angles for both sliding and toppling.

1. Condition for Sliding:
The block begins to slide when the component of gravity acting down the inclined plane exceeds the maximum static frictional force:
$mg \sin\theta \ge f_{\text{max}} = \mu mg \cos\theta$
$\tan\theta \ge \mu$
Given that $\mu = \sqrt{3}$:
$\tan\theta_s = \sqrt{3} \implies \theta_s = 60^\circ$
Thus, sliding would occur at an angle of $\theta_s = 60^\circ$ if the block does not topple earlier.

2. Condition for Toppling:
Let the width of the base along the incline be $b = 10\text{ cm}$ and the height of the block be $h = 15\text{ cm}$. The center of mass is located at a distance of $\frac{h}{2}$ perpendicular to the incline and $\frac{b}{2}$ parallel to the incline from the lower edge.

As $\theta$ increases, the normal force shifts toward the lower edge of the base to provide a counter-torque against the toppling torque produced by the component of gravity down the plane ($mg \sin\theta$).

At the verge of toppling, the normal force acts right at the lowest edge of the block. Taking torque about this lower edge:
$\tau_{\text{toppling}} = (mg \sin\theta) \left(\frac{h}{2}\right)$
$\tau_{\text{restoring}} = (mg \cos\theta) \left(\frac{b}{2}\right)$

For rotational equilibrium:
$(mg \sin\theta)\frac{h}{2} \le (mg \cos\theta)\frac{b}{2}$
$\tan\theta \le \frac{b}{h}$

Hence, the critical angle for toppling is:
$\tan\theta_t = \frac{b}{h} = \frac{10}{15} = \frac{2}{3} \implies \theta_t = \arctan\left(\frac{2}{3}\right) \approx 33.7^\circ$

Conclusion:
Comparing the two critical angles:
$\theta_t \approx 33.7^\circ < \theta_s = 60^\circ$

Since $\theta_t < \theta_s$, as $\theta$ is gradually increased from $0^\circ$, the angle for toppling is reached first. Therefore, the block will remain at rest on the plane up to $\theta = \arctan(2/3)$ and then it will topple without sliding.

Hence, the correct option is B.