A small block of mass of $0.1\ \mathrm{kg}$ lies on a fixed inclined plane $PQ$ which makes an angle $\theta$ with the horizontal. A horizontal force of $1\ \mathrm{N}$ acts on the block through its centre of mass as shown in the figure. The block remains stationary if (take $g = 10\ \mathrm{m/s^2}$)

Step 1: Identify the given parameters.

The mass of the block and the horizontal force acting on it are given.

$$m = 0.1\,\text{kg}$$ $$g = 10\,\text{m/s}^{2}$$ $$F = 1\,\text{N}$$

Step 2: Resolve the forces along the inclined plane.

The component of the weight acting down the incline (towards $P$) is determined by the gravitational force.

$$mg = 0.1 \times 10 = 1\,\text{N}$$ $$mg_{\parallel} = mg\sin\theta = \sin\theta$$

The component of the horizontal force acting up the incline (towards $Q$) is determined by the force $F$.

$$F_{\parallel} = F\cos\theta = \cos\theta$$

Step 3: Analyze the condition for $\theta = 45^\circ$.

Substitute $\theta = 45^\circ$ into the force components.

$$F_{\parallel} = \cos 45^\circ = \frac{1}{\sqrt{2}}\,\text{N}$$ $$mg_{\parallel} = \sin 45^\circ = \frac{1}{\sqrt{2}}\,\text{N}$$

Since the forces along the incline are perfectly balanced, the friction force is zero.

$$f = 0$$

Thus, the block remains stationary.

Step 4: Analyze the condition for $\theta > 45^\circ$.

When the angle of inclination is greater than $45^\circ$, the sine component exceeds the cosine component.

$$\sin\theta > \cos\theta$$ $$mg\sin\theta > F\cos\theta$$

The block has a tendency to slide down the incline towards $P$, meaning the friction force $f$ must act up the incline towards $Q$ to keep it stationary.

Step 5: Analyze the condition for $\theta < 45^\circ$.

When the angle of inclination is less than $45^\circ$, the cosine component exceeds the sine component.

$$\cos\theta > \sin\theta$$ $$F\cos\theta > mg\sin\theta$$

The block has a tendency to slide up the incline towards $Q$, meaning the friction force $f$ must act down the incline towards $P$ to keep it stationary.

Answer: The block remains stationary if $\theta = 45^\circ$ or if $\theta > 45^\circ$ with friction acting up the incline towards $Q$.