JEE Main 202528 JanuaryMorning Shift
- Exam
- JEE Main
- Year
- 2025
- Date
- 28 January
- Shift
- Morning Shift
- Subject
- Physics
- Chapter
- Work, Power and Energy
JEE Main 2025 · 28 January · Morning Shift · Physics · Work, Power and Energy
A bead of mass ' m ' slides without friction on the wall of a...
A bead of mass ' ' slides without friction on the wall of a vertical circular hoop of radius ' ' 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 ' '. 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)

Options
- A
- B
- C
- D
Correct answer
A.
Step-by-step solution
\frac{3 m g R}{2}+\frac{k R^2}{2}=\frac{1}{2} m v^2
or \quad v^2=3 g R+\frac{k R^2}{m}
or \quad v=\sqrt{3 g R+\frac{k R^2}{m}}$