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JEE Main 202623 JanuaryEvening Shift
Exam
JEE Main
Year
2026
Date
23 January
Shift
Evening Shift
Subject
Physics
Chapter
Work, Power and Energy

JEE Main 2026 · 23 January · Evening Shift · Physics · Work, Power and Energy

A bead P sliding on a frictionless semi-circular string (A C B) and...

A bead PP sliding on a frictionless semi-circular string (ACB)(A C B) and it is at point SS at t=0t =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=0t=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
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

Options

  1. AtP<tQ{ }^{t_{P} < t_{Q}}
  2. BtP>tQ{ }^{t_{P}>t_{Q}}
  3. CtP=tQ{ }^{t_{P}=t_{Q}}
  4. DtP>1.25tQt_{P}>1.25 t_{Q}

Correct answer

A. tP<tQ{ }^{t_{P} < t_{Q}}

Step-by-step solution

Horizontal displacement of Q is more than P.
XQ>XPX_Q > X_P
Horizontal component of velocity is same
So tp=xpvt_p = \dfrac{x_p}{v}
tQ=xQvt_Q = \dfrac{x_Q}{v}
tQ>tpt_Q > t_p
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