The figure here shows a heavy box sliding freely on fixed smooth very long incline. A smooth rigid wire PQ is fixed to the box as shown. A bead R (of mass 1 kg) attached to a spring of natural length  $(\sqrt{3}+1)m$  can slide freely along the wire. Initially At the instant shown, the spring is relaxed and oriented perpendicular to the incline while the bead is at rest w.r.t wire.During subsequent motion, the beads comes to rest w.r.t wire when spring has rotated by  ${30}^0$. Taking, $g=10m/s^2$ ,choose the correct alternative(s).

From FBD of box shown in Fig.1
 
Acceleration of box,
$a=\frac{\text{Mg}\sin \theta }{\text{M}}=\text{g}\sin \text{θ}$ 
$\Rightarrow {\text{ a}}_{\text{box}}=\text{g}\sin$ 
$\therefore$  Relative acceleration due to gravity inside box,
  ${\text{g}}^{\text{'}}=\text{g}-\text{a}$
 $=\left(\text{g}\sin \text{θ}\stackrel{\wedge }{i}\text{-gcosθ}\stackrel{\wedge }{j}\right)-g\sin \theta \hat{i}$
 $-g\cos \theta \hat{j}$
So, relative acceleration due to gravity,
 ${\text{g}}^{\text{'}}=\text{g}\cos \text{θ}=\text{g}\cos {60}^0=\frac{\text{g}}{2}=5{\text{ ms}}^{\text{-2}}$
If we rotate the figure of box for simplicity of analysis, we get Fig. 2. Let natural length of spring is I (initially) and spring constant is k.
 
Let extension in spring in final position T be x. By sine rule in  $\Delta RST$ , we get
 $$\begin{gathered} \frac{I+x}{\sin {120}^0}=\frac{I}{\sin {30}^0} \\ \Rightarrow I+x=\sqrt{3}I \\ \Rightarrow \text{ x=}(\sqrt{3}-1)I=(\sqrt{3}-1)(\sqrt{3}+1) \\ \Rightarrow x=2\text{m }\mathrm{...}\text{(i)} \end{gathered}$$
Maximum displacement of the bead is equal to
 $\text{d = RT = I = }(\sqrt{3}+1)\text{m }\left(\text{given}\right)\mathrm{...}\text{ (ii)}$
As, bead comes to rest at T, after release, loss in its potential energy is equal to gain in spring potential energy.
Therefore,  ${\text{mg}}^{\text{'}}\frac{\text{d}}{2}=\frac{1}{2}{\text{kx}}^2$
 $\Rightarrow \text{ k=}\frac{{\text{mg}}^{\text{'}}d}{x^2}=\frac{1\times 5\times (\sqrt{3}+1)}{4}$
   $=\frac{5(\sqrt{3}+1)}{4}N/m$                         
[using Eqs. (i) and (ii)]
From FBD of bead in position T as shown Fig.2, in direction perpendicular to the wire, we get
 $N+\frac{kx}{2}=\frac{\sqrt{3}m{g}^{\text{'}}}{2}\Rightarrow \text{N}=\frac{\sqrt{3}m{g}^{\text{'}}-kx}{2}$
 $=\frac{\sqrt{3}\times 1\times 5-\frac{5(\sqrt{3}+1)}{4}\times 2}{2}$
 $=\frac{5(\sqrt{3}-1)}{4}\text{N}$