
Work done by gravity =mg(2R−Rcos60∘)=23mgR Work done by spring =−21k(02−R2)=21kR2 Net work = change in kinetic energy i.e. 23mgR+2kR2=21mv2 or v2=3gR+mkR2 or v=3gR+mkR2
JEE Main 2025 — Physics Mechanics
A bead of mass ' m ' slides without friction on the wall of a vertical circular hoop of radius ' R ' as shown in 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 top of the hoop with (negligible) zero initial speed, velocity of bead, when the length of spring becomes ' R ', would be (spring constant is ' k ', g is accleration due to gravity) 
Held on 28 Jan 2025 · Verified 6 Jul 2026.
3Rg+ mkR2
2gR+ mkR2
2Rg+ mkR2
2Rg+ m4kR2
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