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Physics · Work, Power & Energy

JEE Main 2025 — 28 January, Morning Shift — Question 58

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 acceleration due to gravity) IMAGES

Question figure
  1. Option A:

    2gR+kR2m2\sqrt{gR+\frac{k{{R}^{2}}}{m}}

  2. Option B:

    2Rg+4kR2  ⁣ ⁣  ⁣ ⁣ m\sqrt{2\text{Rg}+\frac{4\text{k}{{\text{R}}^{2}}}{\text{ }\!\!~\!\!\text{ m}}}

  3. Option C:

    2Rg+kR2m\sqrt{2Rg+\frac{\text{k}{{\text{R}}^{2}}}{m}}

  4. Option D:

    3Rg+kR2m\sqrt{3Rg+\frac{k{{R}^{2}}}{m}}

    Correct

Answer: D

Step-by-step solution

Work energy theorem Mg(R+Rcos60)+12k(R2−02)=12mv2\text{Mg}\left( \text{R}+\text{Rcos}60 \right)+\frac{1}{2}\text{k}\left( {{\text{R}}^{2}}-{{0}^{2}} \right)=\frac{1}{2}\text{m}{{\text{v}}^{2}} Mg3R2+KR22=12mv2\text{Mg}\frac{3\text{R}}{2}+\frac{\text{K}{{\text{R}}^{2}}}{2}=\frac{1}{2}\text{m}{{\text{v}}^{2}} V=3gR+KR2mV=\sqrt{3gR+\frac{K{{R}^{2}}}{m}}

Solution figure

Answer key and solution verified before publishing.

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Exam
JEE Main 2025
Paper
28 January, Morning Shift
Subject
Physics
Chapter
Work, Power & Energy
Topic
Applications of Conservation of Mechanical Energy

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