JEE Mains · Physics · STD 11 - Work,energy,power and collision
- JEE Mains 2025
- Hard
- MCQ
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)

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Answer & Solution
Correct Answer
(A) \(\sqrt{3 \mathrm{Rg}+\frac{\mathrm{kR}^2}{\mathrm{~m}}}\)
Step-by-step Solution
Detailed explanation

\(\begin{aligned} \text { Work done by gravity } & =m g\left(2 R-R \cos 60^{\circ}\right) \\ & =\frac{3 m g R}{2}\end{aligned}\)
\(\begin{aligned} \text { Work done by spring } & =-\frac{1}{2} k\left(0^2-R^2\right) \\ & =\frac{1}{2} k R^2\end{aligned}\)
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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