JEE Main · 2025 · 28 Jan 2025 · Shift 1 · Work, Power & Energy

Place the center of the hoop at O and let P be the fixed point at the bottom where the spring is anchored, so OP = R. The bead starts at the point diametrically opposite P, i.e. at the top of the hoop, where the chord to P has length 2R. Since the spring's natural length is R, it starts stretched by an amount 2R - R = R.

When the spring length becomes R again (natural length), the chord from the bead to P equals R. In the triangle formed by O, P and the bead, two sides (OP and O-bead) equal R, and now the third side (the chord) also equals R, so this is an equilateral triangle: the bead has swept through 60° as seen from O.

Taking O as origin with P at the bottom, the top of the hoop is at height R above O, while the bead's new position is at height $R\cos 60^{\circ }$ below O, i.e. at $-\frac{R}{2}$. So the bead has descended a total height


Applying the work-energy theorem to the bead (gravity does positive work as it falls, and the spring's stored elastic energy is released as it relaxes from extension R to zero):