Horizontal displacement of Q is more than P.
XQ>XP
Horizontal component of velocity is same
So tp=vxp
tQ=vxQ
tQ>tp
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 tP and tQ be the respective times taken by beads P and Q to reach the point B, then the relation between tP and tQ is

Held on 23 Jan 2026 · Verified 6 Jul 2026.
tP<tQ
tP>tQ
tP=tQ
tP>1.25tQ
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