JEE Main 2026 — Jan 23, Shift 2 — Physics solution

JEE Main2026Jan 23, Shift 2Physics

A bead $P$ sliding on a frictionless semi-circular string ( $A C B$ ) 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 $A B$ , 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

A
$t_{P} > t_{Q}$
B
$t_{p} < t_{Q}$
C
$t_{P} > 1.25 t_{Q}$
D
$t_{p} = t_{Q}$

Solution

Bead $P$ slides on the curved string from $S$ down to $C$ and back up to $B$ . By energy conservation, gravity supplies kinetic energy as it descends and removes it on the way up. Crucially, the horizontal displacement of $P$ from $S$ to $B$ equals $X_{P}$ (a chord projection), while bead $Q$ travels the full horizontal distance $X_{Q} = A B$ from $A$ to $B$ at constant speed $v$ .

So $X_{Q} > X_{P}$ .

Bead $P$ on the curve gains speed (above $v$ ) over part of its path due to the descent, so its average horizontal speed exceeds $v$ . But the path's horizontal projection is shorter:

$t_{P} = \frac{X _{P}}{v ˉ _{x}^{(P)}}$ where $\overset{v}{ˉ}_{x}^{(P)} \geq v$ , and $t_{Q} = \frac{X _{Q}}{v}$ with $X_{Q} > X_{P}$ .

Combining: $t_{P} < t_{Q}$ .

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