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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 ' mm ' slides without friction on the wall of a vertical circular hoop of radius ' RR ' 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 ' RR '. 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) A bead of mass ' m ' slides without friction on the wall of a vertical circular hoop of radius ' R ' as shown in

Options

  1. A3Rg+kR2 m\sqrt{3 \mathrm{Rg}+\frac{\mathrm{kR}^2}{\mathrm{~m}}}
  2. B2gR+kR2 m2 \sqrt{\mathrm{gR}+\frac{\mathrm{kR}^2}{\mathrm{~m}}}
  3. C2Rg+kR2 m\sqrt{2 \mathrm{Rg}+\frac{\mathrm{kR}^2}{\mathrm{~m}}}
  4. D2Rg+4kR2 m\sqrt{2 \mathrm{Rg}+\frac{4 \mathrm{kR}^2}{\mathrm{~m}}}

Correct answer

A. 3Rg+kR2 m\sqrt{3 \mathrm{Rg}+\frac{\mathrm{kR}^2}{\mathrm{~m}}}

Step-by-step solution

 Work done by gravity =mg(2R−Rcos⁡60∘)=3mgR2\begin{aligned} \text{ Work done by gravity } & =m g\left(2 R-R \cos 60^{\circ}\right) \\ & =\frac{3 m g R}{2}\end{aligned} Work done by spring =−12k(02−R2)=12kR2\begin{aligned} \text{ Work done by spring } & =-\frac{1}{2} k\left(0^2-R^2\right) \\ & =\frac{1}{2} k R^2\end{aligned}Network=changeinkineticenergyi.e.Net work = change in kinetic energy i.e.\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}}$
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