Physics · Motion in Plane

JEE Main 2026 — 23 January, Evening Shift — Question 32

A bead PP sliding on a frictionless semi-circular string ( ACBA C B ) and it is at point SS at t=0\mathrm{t}=0 and at this instant the horizontal component of its velocity is vv. Another bead QQ of the same mass as PP is ejected from point AA at t=0\mathrm{t}=0 along the horizontal string ABA B, with the speed vv, friction between the beads and the respective strings may be neglected in both cases. Let tpt_{p} and tQt_{Q} be the respective times taken by beads PP and QQ to reach the point BB, then the relation between tpt_{p} and tQt_{Q} is

Question figure
  1. Option A:

    tP>tQt_{P}>t_{Q}

  2. Option B:

    tp<tQt_{p} < t_{Q}

    Correct
  3. Option C:

    tp>1.25tQt_{p} > 1.25 t_{Q}

  4. Option D:

    tp=tQt_{p}=t_{Q}

Answer: B

Step-by-step solution

Horizontal displacement of Q is more then P .

XQ>XPX_{Q}>X_{P}

Horizontal component of velocity is same So tp=Xpv\mathrm{t}_{\mathrm{p}}=\frac{\mathrm{X}_{\mathrm{p}}}{\mathrm{v}}

tQ=xQvt_{Q}=\frac{x_{Q}}{v}

tQ>tp\mathrm{t}_{\mathrm{Q}}>\mathrm{t}_{\mathrm{p}}

Answer key and solution verified before publishing.

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Exam
JEE Main 2026
Subject
Physics
Chapter
Motion in Plane
Topic
Oblique and Horizontal Projectile Motion
A bead P sliding on a frictionless semi-circular string ( A C B ) and… | JEE Main 2026 PYQ with Solution · DhiX AI