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

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

A small bob A of mass m is attached to a massless rigid rod of...

A small bob AA of mass mm is attached to a massless rigid rod of length 1 m pivoted at point PP and kept at an angle of 60∘60^{\circ} with vertical as shown in figure. At distance of 1 m below point PP, an identical bob BB is kept at rest on a smooth horizontal surface that extends to a circular track of radius RR as shown in figure. If bob BB just manages to complete the circular path of radius RR upto a point QQ after being hit elastically by bob⁡A\operatorname{bob} A, then radius RR is ____\_\_\_\_ m.
A small bob A of mass m is attached to a massless rigid rod of length 1 m pivoted at point P and kept at an angle of

Options

  1. A15\frac{1}{5}
  2. B2−35\frac{2-\sqrt{3}}{5}
  3. C35\frac{3}{5}
  4. D2+35\frac{2+\sqrt{3}}{5}

Correct answer

A. 15\frac{1}{5}

Step-by-step solution

Bob A swings from 60°60° to vertical. Height drop h=l(1−cos⁡60°)=1×0.5=0.5h = l(1-\cos 60°) = 1 \times 0.5 = 0.5 m.
Velocity of A at bottom: vA=2gh=2×10×0.5=10v_A = \sqrt{2gh} = \sqrt{2 \times 10 \times 0.5} = \sqrt{10} m/s.
Elastic collision between equal masses: B gets vB=10v_B = \sqrt{10} m/s, A stops.
For B to just complete circular loop: vB2=5gRv_B^2 = 5gR.
10=5×10×R⇒R=1510 = 5 \times 10 \times R \Rightarrow R = \frac{1}{5} m.
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