MCQMediumJEE 2026Projectile Motion

JEE Physics 2026 Question with Solution

A bead P sliding on a frictionless semi-circular string... bead Q ejected... relation between tPt_P and tQt_Q is

A missing physics diagram comparing bead P sliding on a frictionless semicircular string and bead Q being ejected, with times t_P and t_Q to be compared.
  • A

    tP>tQt_P > t_Q

  • B

    tP>1.25tQt_P > 1.25 t_Q

  • C

    tP=tQt_P = t_Q

  • D

    tP<tQt_P < t_Q

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given: Bead P slides on a frictionless semicircular string, while bead Q is ejected. We need to compare the times tPt_P and tQt_Q.

Find: The correct relation between tPt_P and tQt_Q.

From the extracted solution: bead Q moves with constant speed vv.

Bead P accelerates due to gravity along the curved path, so its average speed becomes greater than vv.

Although bead P travels along a longer path, the increase in speed makes its travel time smaller.

Therefore, tP<tQt_P < t_Q. The correct option is D.

Common mistakes

  • Assuming the shorter geometric path must always take less time. This is wrong because time depends on both path length and speed. Here bead P gains speed due to gravity, so compare travel time, not only distance.

  • Treating both beads as moving with the same constant speed. This is wrong because only bead Q moves with constant speed vv, while bead P accelerates on the curved path. Use the change in speed for bead P.

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