One end of a horizontal uniform beam of weight W and length L is...

$∵$ Rod is in equilibrium, net torque on the rod about $O$ will be zero.

$W\left(\dfrac{l}{2}\right) + \alphaW\left(l\right) = \dfrac{T}{\sqrt{2}}l$

$⇒ \dfrac{W}{2} + \alphaW = \dfrac{T}{\sqrt{2}} ⇒ T = \sqrt{2}W\left(\alpha + 0 . 5\right) . . . \left(1\right)$

Now checking options-

(A)

Force balance on rod,

${N}_{v} + \dfrac{T}{\sqrt{2}} = W + \alphaW$

$⇒ {N}_{v} = \left(W + \alphaW\right) - \left(\dfrac{W}{2} + \alphaW\right)$

$⇒ {N}_{v} = \dfrac{W}{2}$

Hence, option (A) is correct.

(B) Write horizontal force equation,

${N}_{H} = \dfrac{T}{\sqrt{2}} ⇒ {N}_{H} = W\left(\alpha + 0 . 5\right)$

If $\alpha = 0 . 5$, then,

${N}_{H} = W$ and $T = \sqrt{2}W$

Hence, option (B) is correct.

(D) $∴$ Maximum Tension of string $= 2\sqrt{2}W$

So, to break.

$T > 2\sqrt{2}W ⇒ \sqrt{2}W\left(\alpha + 0 . 5\right) > 2\sqrt{2}W ⇒ \alpha > 1 . 5$

So, string will break.

option (D) correct.