JEE Main · 2026 · 24 Jan 2026 · Shift 2 · Work, Power & Energy

At the position shown the thread makes $30^{\circ }$ with the horizontal, hence $60^{\circ }$ with the upward vertical. With the tension zero there, the inward (centripetal) force is supplied by the radial component of gravity: $mg\cos 60^{\circ }=\frac{m{v}^2}{r}$, giving $v^2=\frac{gr}{2}$. That point stands a height $r+r\cos 60^{\circ }=\frac{3r}{2}$ above the lowest point $A$. Energy conservation from $A$ gives $\frac{1}{2}{u}^2=\frac{1}{2}{v}^2+g\cdot \frac{3r}{2}$, so $u^2=\frac{gr}{2}+3gr=\frac{7gr}{2}$. The speed at $A$ is $\sqrt{\frac{7}{2}gr}$.