A bead P sliding on a frictionless semi-circular string (ACB) and it… | JEE Main 2026 (23 Jan, Shift 2)

Q4JEE Main 2026 · 23 JAN · SHIFT 2Difficult

Conservation of Energy

A bead $P$ sliding on a frictionless semi-circular string $(ACB)$ and it is at point $S$ at $t = 0$ and at this instant the horizontal component of its velocity is $v$ . Another bead $Q$ of the same mass as $P$ is ejected from point $A$ at $t = 0$ along the horizontal string $AB$ , with the speed $v$ , friction between the beads and the respective strings may be neglected in both cases. Let $t_P$ and $t_Q$ be the respective times taken by beads $P$ and $Q$ to reach the point $B$ , then the relation between $t_P$ and $t_Q$ is

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Answer: (a)

Bead Q has constant speed $v$ along AB: $t_Q = \frac{2R}{v}$ .
At S the radius is at $45^\circ$ , so the velocity is at $45^\circ$ too: speed of P $= \frac{v}{\cos 45^\circ} = \sqrt{2}v$ .
Path of P is arc SCB $= \frac{3\pi R}{4} \approx 2.36R$ ; at speed $\sqrt{2}v$ this takes about $\frac{2.36R}{1.41v} \approx 1.67\frac{R}{v}$ .
P speeds up while dipping to C, so its time stays below $2R/v$ (a full calculation gives $t_P \approx 0.83$ to $0.95\,t_Q$ ).
So the answer is (a) $t_P < t_Q$ .

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