Question

A bead of mass $m$ is attached to one end of a spring of natural length $R$ and spring constant $K=\frac{(\sqrt{3}+1)mg}{R}$. The other end of the spring is fixed at a point $A$ on a smooth vertical ring of radius $R$ as shown in $Fig.6.301$ The normal reaction at $B$ just after it is released to move is

• A. $mg/2$
• B. $\sqrt{3}mg$
• C. $3\sqrt{3}mg$
• D. $\frac{3\sqrt{3}mg}{2}$

Answer 1

The correct option is D $\frac{3\sqrt{3}mg}{2}$

From figure length $AB=\sqrt{R^2+R^2+2R\cos 60}=\sqrt{3}R$

(We know that the resultant of two vectors$\overrightarrow{A}and\overrightarrow{B}$ having angle $\theta$ between them $=\sqrt{A^2+B^2+2AB\cos \theta }$

Natural length $=R\left(given\right)$

$Extension=\sqrt{3}R-R=(\sqrt{3}-1)R$

$⟹F_s=K\times extension=\frac{(\sqrt{3}+1)mg}{R}\times (\sqrt{3}-1)R=2mg$

From last figure it is clear that $N=F_s\cos 30+mg\cos 30$

$⟹N=3mg\cos 30=\frac{3\sqrt{3}}{2}mg$