A block of base $10\text{ cm} \times 10\text{ cm}$ and height $15\text{ cm}$ is kept on an inclined plane. The coefficient of friction between them is $\sqrt{3}$. The inclination $\theta$ of this inclined plane from the horizontal plane is gradually increased from $0^\circ$. Then

Step 1: Identify the given parameters of the block and the inclined plane.

The dimensions of the block and the coefficient of friction are:

$$b = 10\,\text{cm}$$ $$h = 15\,\text{cm}$$ $$\mu = \sqrt{3}$$

Step 2: Determine the critical angle for sliding.

For the block to slide, the component of gravity along the incline must exceed the maximum static friction force.

$$mg\sin\theta > \mu mg\cos\theta$$ $$\tan\theta > \mu$$

The critical angle for sliding $\theta_s$ is given by:

$$\tan\theta_s = \sqrt{3}$$ $$\theta_s = 60^\circ$$

Step 3: Determine the critical angle for toppling.

The block will topple about its lower edge when the line of action of the gravitational force passes outside its base. This transition occurs when the torque about the lower edge becomes zero.

$$h\sin\theta = b\cos\theta$$ $$\tan\theta_t = \frac{b}{h}$$

Substitute the given dimensions of the block:

$$\tan\theta_t = \frac{10}{15} = \frac{2}{3}$$ $$\theta_t = \tan^{-1}\left(\frac{2}{3}\right) \approx 33.7^\circ$$

Step 4: Compare the critical angles to determine the nature of motion.

Compare the two critical angles:

$$\theta_t \approx 33.7^\circ < \theta_s = 60^\circ$$

Since the critical angle for toppling is smaller than the critical angle for sliding, the block will topple before it can slide.

Answer: The block will topple at an angle of $\theta = \tan^{-1}\left(\dfrac{2}{3}\right) \approx 33.7^\circ$ before it starts sliding.