A bead of mass ' $m$ ' slides without friction on the wall of a vertical circular hoop of radius ' $R$ ' as…
- $\sqrt{3 \mathrm{Rg}+\frac{\mathrm{kR}^2}{\mathrm{~m}}}$
- $2 \sqrt{\mathrm{gR}+\frac{\mathrm{kR}^2}{\mathrm{~m}}}$
- $\sqrt{2 \mathrm{Rg}+\frac{\mathrm{kR}^2}{\mathrm{~m}}}$
- $\sqrt{2 \mathrm{Rg}+\frac{4 \mathrm{kR}^2}{\mathrm{~m}}}$
Solution
$\begin{aligned} \text { Work done by gravity } & =m g\left(2 R-R \cos 60^{\circ}\right) \\ & =\frac{3 m g R}{2}\end{aligned}$
$\begin{aligned} \text { Work done by spring } & =-\frac{1}{2} k\left(0^2-R^2\right) \\ & =\frac{1}{2} k R^2\end{aligned}$
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 $\quad v=\sqrt{3 g R+\frac{k R^2}{m}}$
Asked in: JEE Main 2025 (28 Jan Shift 1)