A bead of mass $m$ slides without friction on the wall of a vertical circular hoop of radius $R$ as shown in the figure. The bead moves under the combined action of gravity and a massless spring ($k$) attached to the bottom of the hoop. The equilibrium length of the spring is $R$. If the bead is released from the top of the hoop with (negligible) zero initial speed, the velocity of the bead when the length of the spring becomes $R$ would be (spring constant is $k$, $g$ is acceleration due to gravity)
Answer: (A) $\sqrt{3Rg + \dfrac{kR^2}{m}}$
At the top the spring stretches across the diameter: length $2R$, extension $R$.
When the spring length is $R$ (natural length), the bead is at chord distance $R$ from the bottom point. For a chord of length $c$ from the lowest point of a circle, the height is $h = \dfrac{c^2}{2R} = \dfrac{R}{2}$.
So the bead falls from height $2R$ to $R/2$, a drop of $\tfrac{3R}{2}$, and the spring gives up all its energy:
$$\frac{1}{2}mv^2 = mg\cdot\frac{3R}{2} + \frac{1}{2}kR^2 \Rightarrow v = \sqrt{3Rg + \frac{kR^2}{m}}$$
Solution by Sreeraj P, M.Sc Physics