1. Given data

• Mass of block: $m = 0.1\,\text{kg}$
• Gravitational acceleration: $g = 10\,\text{m/s}^2$
• Hence weight: $mg = 0.1 \times 10 = 1\,\text{N}$
• A horizontal force of magnitude $1\,\text{N}$ acts on the block.

So, both the weight and the applied horizontal force have magnitude $1\,\text{N}$.

2. Resolve forces along the incline

Let the incline rise from $P$ to $Q$, so "up the plane" means towards $Q$.

(i) Component of weight along plane

Weight acts vertically downward, so its component along the plane is $mg\sin\theta = 1\cdot \sin\theta = \sin\theta$ This acts down the plane, i.e. towards $P$.

(ii) Component of horizontal force along plane

The horizontal force is towards the right as shown, and its component along the incline is $F\cos\theta = 1\cdot \cos\theta = \cos\theta$ This acts up the plane, i.e. towards $Q$.

3. Compare the two components

Net force component along plane (without friction) is $\cos\theta - \sin\theta$

• If $\cos\theta > \sin\theta$, tendency is up the plane (towards $Q$), so friction must act down the plane (towards $P$).
• If $\cos\theta < \sin\theta$, tendency is down the plane (towards $P$), so friction must act up the plane (towards $Q$).
• If $\cos\theta = \sin\theta$, no friction is needed.

Now, $\cos\theta = \sin\theta \iff \theta = 45^\circ$

So:

• For $\theta = 45^\circ$, block can remain stationary without friction along the plane.
• For $\theta > 45^\circ$, we have $\sin\theta > \cos\theta$, so tendency is down the plane, hence friction acts up the plane, i.e. towards $Q$.
• For $\theta < 45^\circ$, we have $\cos\theta > \sin\theta$, so tendency is up the plane, hence friction acts down the plane, i.e. towards $P$.

4. Check each option

Option A: $\theta = 45^\circ$

At $\theta = 45^\circ$, $\sin 45^\circ = \cos 45^\circ$ So the two along-plane components cancel. The block can remain stationary.

✅ A is correct

Option B: $\theta > 45^\circ$ and friction acts towards $P$

For $\theta > 45^\circ$, the block tends to move towards $P$ due to larger $mg\sin\theta$. Therefore friction must oppose this and act towards $Q$, not $P$.

❌ B is incorrect

Option C: $\theta > 45^\circ$ and friction acts towards $Q$

This matches the correct direction of friction.

✅ C is correct

Option D: $\theta < 45^\circ$ and friction acts towards $Q$

For $\theta < 45^\circ$, the block tends to move towards $Q$, so friction should act towards $P$, not $Q$.

❌ D is incorrect

5. Final answer

The correct options are: $\boxed{A,\ C}$

This matches the stored correct answer.