A rod AB is fixed between a vertical wall and a horizontal surface. A smooth bead of mass  $(\sqrt{3}-1)kg$  constrained to move along the rod is released in the position shown in the figure, for which the spring is vertical and relaxed. The natural length of the spring is  $(2+2\sqrt{3})$ m. Rod makes an angle of  ${30}^o$  with the horizontal. The bead again comes to rest when spring makes an angle of  ${30}^o$ with the vertical. The constant of spring is $\frac{k}{2}$ then value of  k  is 

Assuming  $I^{\text{'}}$   to be natural length of spring. Initail and final positions of the spring and bead system during its motion over incline are as shown in figure
 
Let s is displacement of bead and $I^{\text{'}}$  is final spring’s length.
By sine rule in triangle EFC , 
$\frac{I^{\text{'}}}{\sin {120}^0}=\frac{s}{\sin {30}^0}=\frac{I}{\sin {30}^0}$ 
$\Rightarrow I^{\text{'}}=\sqrt{3}I$    and  $s=I\mathrm{...}\text{(i)}$
From  $\Delta EDC,ED=s{\text{ sin30}}^0=\frac{I}{2}\mathrm{...}\text{(ii)}$
So, by conservation of energy for motion of spring-bead system.
$\Rightarrow \frac{1}{2}k{(\sqrt{3}I-I)}^2=mg\frac{I}{2}$       
 [using Eqs. (i) and (ii)]
$\Rightarrow k=\frac{mg}{{(\sqrt{3}-1)}^{2}I}$ 
  
$=\frac{(\sqrt{3}-1)\times 10}{{(\sqrt{3}-1)}^2(\sqrt{3}+1)\times 2}=2.5N/m$