Finally the spring is unstretched, If the plane is mow gradually lifted from $\theta=0^{\circ}$ to $\theta=90^{\circ}$ then the graph showing extension in the spring(x) versus angle $(\theta)$ is

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$x=0$ till $mg \sin \theta < \mu mg \cos\theta$

ie no extension takes place is less than the frictional force.

gradually x will increase at angle $\theta > \tan ^{-1} \mu$

$kx+\mu mg \cos \theta =mg \sin \theta$

$x=\large\frac{mg \sin \theta-\mu mg \cos \theta}{k}$

where k is spring constant

Hence a is the correct answer.

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