The uniform link shown in the following figure has a mass of 12 kg. If the spring is un-stretched at θ = 0$$\degree$$​, determine the angle θ for stable equilibrium. Neglect the surface friction and assume g = 10 $$m/s^2$$)

Correct option is B

​$$\text{The spring is unstretched at } \theta = 0^\circ. \text{ When the rod makes an angle } \theta \text{ with the vertical,} \\\text{the end at A moves down by } x = l(1 - \cos \theta), \text{ so spring force is } F_s = kx. \\[10pt]\textbf{Considering Equilibrium:} \\[4pt]\text{Moments about A: } \quad N(l \sin \theta) = mg\left(\frac{l}{2}\sin \theta\right) \Rightarrow N = \frac{mg}{2}. \\[6pt]\text{Vertical forces: } \quad N + F_s = mg \Rightarrow F_s = \frac{mg}{2}. \\[10pt]kx = \frac{mg}{2} \Rightarrow x = \frac{mg}{2k} = \frac{12 \times 10}{2 \times 200} = 0.3 \text{ m.} \\[8pt]x = l(1 - \cos \theta) = 0.6(1 - \cos \theta) = 0.3 \Rightarrow \cos \theta = \frac{1}{2} \Rightarrow \theta = 60^\circ.$$​​