A wheel of radius $R$ and mass $M$ is placed at the bottom of a fixed step of height $R$ as shown in the figure. A constant force is continuously applied on the surface of the wheel so that it just climbs the step without slipping. Consider the torque $\tau$ about an axis normal to the plane of the paper passing through the point $Q$. Which of the following options is/are correct?

Step 1: Define the torque about the pivot point $Q$.

The wheel pivots about the corner of the step at point $Q$, and the torque is determined by the perpendicular distance from $Q$ to the line of action of the force.

$$\vec{\tau}_Q = \vec{r} \times \vec{F}$$ $$|\tau_Q| = F \cdot d$$

Since the force is continuously applied at a fixed point on the wheel's surface, the lever arm $d$ remains constant as the wheel rotates about $Q$.

Step 2: Analyze a tangential force applied at point $P$.

The line of action of a tangential force at $P$ is perpendicular to the diameter $PQ$.

$$d = QP = 2R$$ $$\tau = 2FR$$

Since the torque is constant and does not decrease, Option (1) is incorrect.

Step 3: Analyze a normal force applied at point $X$.

The line of action of a normal force at $X$ passes through the center $C$ of the wheel.

$$d = R\sin\alpha$$ $$\tau = FR\sin\alpha$$

Since $\alpha$ is the constant angle between $QC$ and $CX$, the torque is constant and non-zero. Thus, Option (2) is correct.

Step 4: Analyze a normal force applied at point $P$.

The line of action of a normal force at $P$ is along the diameter $PQ$ and passes directly through $Q$.

$$d = 0$$ $$\tau = 0$$

Thus, Option (3) is correct.

Step 5: Analyze a tangential force applied at point $S$.

The perpendicular distance from $Q$ to the horizontal tangential force at the top point $S$ is $R$.

$$d = R$$ $$\tau_F = FR$$

The restoring torque due to gravity about $Q$ decreases as the climb angle $\theta$ increases.

$$\tau_g = MgR\cos\theta$$

For a sufficiently large force, the wheel climbs the step, making Option (4) incorrect.

Answer: The correct options are (2) and (3).