One end of a horizontal uniform beam of weight $W$ and length $L$ is hinged on a vertical wall at point $O$ and its other end is supported by a light inextensible rope. The other end of the rope is fixed at point $Q$, at a height $L$ above the hinge at point $O$. A block of weight $\alpha W$ is attached at the point $P$ of the beam, as shown in the figure (not to scale). The rope can sustain a maximum tension of $(2\sqrt{2})\,W$. Which of the following statement(s) is(are) correct?

Step 1: Determine the angle of the rope.

Since $OQ = OP = L$ and $\angle QOP = 90^\circ$, the triangle $\triangle OPQ$ is a right-angled isosceles triangle.

$$\theta = 45^\circ$$

Step 2: Resolve the tension force into components.

The horizontal and vertical components of the tension $T$ acting at point $P$ are determined by the angle.

$$T_x = T \cos 45^\circ = \frac{T}{\sqrt{2}}$$ $$T_y = T \sin 45^\circ = \frac{T}{\sqrt{2}}$$

Step 3: Apply rotational equilibrium about the hinge $O$.

The sum of the torques about point $O$ must equal zero.

$$T_y \cdot L - W \cdot \frac{L}{2} - \alpha W \cdot L = 0$$ $$\frac{T}{\sqrt{2}} L = W L \left(\frac{1}{2} + \alpha\right)$$ $$T = \frac{W(1 + 2\alpha)}{\sqrt{2}}$$

Step 4: Find the horizontal reaction force $R_{Ox}$ at the hinge.

From horizontal translational equilibrium, the horizontal reaction force equals the horizontal component of tension.

$$R_{Ox} = T_x = \frac{T}{\sqrt{2}}$$ $$R_{Ox} = W \left(\frac{1}{2} + \alpha\right)$$

Step 5: Find the vertical reaction force $R_{Oy}$ at the hinge.

From vertical translational equilibrium, the sum of vertical forces must equal zero.

$$R_{Oy} + T_y = W + \alpha W$$ $$R_{Oy} = W(1 + \alpha) - \frac{T}{\sqrt{2}}$$ $$R_{Oy} = \frac{W}{2}$$

Step 6: Analyze Statement 1.

The vertical component of the reaction force at $O$ is independent of $\alpha$.

$$R_{Oy} = \frac{W}{2}$$

Since $R_{Oy}$ does not depend on $\alpha$, this statement is correct.

Step 7: Analyze Statement 2.

The horizontal component of the reaction force at $O$ is $W$ when $\alpha = 0.5$.

$$R_{Ox}(0.5) = W \left(\frac{1}{2} + 0.5\right) = W$$

This statement is correct.

Step 8: Analyze Statement 3.

The tension in the rope is $2W$ when $\alpha = 0.5$.

$$T(0.5) = \frac{W(1 + 2(0.5))}{\sqrt{2}} = \sqrt{2}W$$

Since $\sqrt{2}W \neq 2W$, this statement is incorrect.

Step 9: Analyze Statement 4.

The rope breaks if $\alpha > 1.5$ given the maximum tension $T_{\max} = 2\sqrt{2}W$.

$$\frac{W(1 + 2\alpha)}{\sqrt{2}} = 2\sqrt{2}W$$ $$1 + 2\alpha = 4$$ $$\alpha = 1.5$$

Since tension increases with $\alpha$, the rope breaks for $\alpha > 1.5$, making this statement correct.

Answer: Statements (1), (2), and (4) are correct.