A bead of mass m can freely slide down the fixed inclined rod without friction. It is connected to a point P on the horizontal surface with a light spring of spring constant k. The bead is initially released from rest and the spring is initially unstressed and vertical. The bead just stops at the bottom of the inclined rod. Find the angle which the inclined rod makes with horizontal.

Applying conservation of energy
Loss in gravitational Potential Energy = gain in spring Potential Energy.
$\Rightarrow mgh=\frac{1}{2}k{x}^2\mathrm{.....}\left(1\right)$

From diagram

$\cot \alpha =\frac{PO}{h}$

$\Rightarrow PO=h\cot \alpha$

$\text{Elongation},x=h\cot \alpha -h$

In this case, the spring will have an elongation but not compression because in case of compression, the particle will not come to rest

$\begin{array}{l}\left(1\right)\Rightarrow mgh=\frac{1}{2}k(h\cot \alpha -h{)}^2\\ \text{ or }(\cot \alpha -1)=\sqrt{\frac{2mg}{kh}}\\ \mathrm{or}(\cot \alpha )=1+\sqrt{\frac{2mg}{kh}}\\ \therefore \alpha ={\cot }^{-1}\left(1+\sqrt{\frac{2mg}{kh}}\right)\end{array}$

Therefore, the correct answer is (A).