1. Set up the geometry

The beam is horizontal, hinged at $O$, and has length $L$.

The rope joins the free end of the beam to point $Q$, which is at height $L$ above $O$. So the rope makes a $45^\circ$ angle with the horizontal.

Hence, if the rope tension is $T$, then its components at the beam end are: $T_x = -\frac{T}{\sqrt{2}}, \qquad T_y = \frac{T}{\sqrt{2}}$ (leftward and upward respectively).

The beam has weight $W$ acting at its center, i.e. at distance $L/2$ from $O$. A block of weight $\alpha W$ is attached at the end point $P$ of the beam (as implied by the figure/text), i.e. at distance $L$ from $O$.

Let hinge reaction at $O$ have components $R_x, R_y$.

2. Torque equilibrium about hinge $O$

Taking anticlockwise torque as positive:

• Upward component of tension at beam end gives torque: $\frac{T}{\sqrt{2}}\cdot L$
• Weight of beam gives clockwise torque: $W\cdot \frac{L}{2}$
• Weight of block gives clockwise torque: $\alpha W\cdot L$

Equilibrium gives $\frac{T}{\sqrt{2}}L - W\frac{L}{2} - \alpha WL = 0$

Cancel $L$: $\frac{T}{\sqrt{2}} = \left(\alpha + \frac12\right)W$

So $T = \sqrt{2}\left(\alpha + \frac12\right)W$

3. Force equilibrium in vertical direction

$R_y + \frac{T}{\sqrt{2}} - W - \alpha W = 0$

Substitute $\frac{T}{\sqrt{2}} = \left(\alpha + \frac12\right)W$

Then $R_y + \left(\alpha + \frac12\right)W - (1+\alpha)W = 0$ $R_y - \frac{W}{2} = 0$ $R_y = \frac{W}{2}$

So the vertical reaction is independent of $\alpha$.

Hence A is correct.

4. Force equilibrium in horizontal direction

Only two horizontal forces act:

• hinge reaction $R_x$ to the right,
• rope component $T/\sqrt{2}$ to the left.

Thus $R_x = \frac{T}{\sqrt{2}} = \left(\alpha + \frac12\right)W$

For $\alpha = 0.5$, $R_x = \left(0.5 + 0.5\right)W = W$

Hence B is correct.

5. Check tension for $\alpha = 0.5$

Using $T = \sqrt{2}\left(\alpha + \frac12\right)W$

For $\alpha=0.5$, $T = \sqrt{2}(1)W = \sqrt{2}W$

This is not equal to $2W$.

Hence C is incorrect.

6. Condition for rope to break

Maximum sustainable tension: $T_{\max} = 2\sqrt{2}\,W$

The rope breaks when $T > 2\sqrt{2}W$

Using $T = \sqrt{2}\left(\alpha + \frac12\right)W$

So $\sqrt{2}\left(\alpha + \frac12\right)W > 2\sqrt{2}W$

Cancel $\sqrt{2}W$: $\alpha + \frac12 > 2$ $\alpha > \frac32$

Thus the rope breaks if $\alpha > 1.5$

Hence D is correct.

7. Final answer

Correct statements are: $\boxed{A,\ B,\ D}$

This matches the stored correct answer.