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JEE Main 202622 JanuaryEvening Shift
Exam
JEE Main
Year
2026
Date
22 January
Shift
Evening Shift
Subject
Physics
Chapter
Rotational Motion

JEE Main 2026 · 22 January · Evening Shift · Physics · Rotational Motion

A uniform bar of length 12 cm and mass 20 m lies on a smooth...

A uniform bar of length 12 cm and mass 20m20 m lies on a smooth horizontal table. Two point masses mm and 2m2 m are moving in opposite directions with same speed of vv and in the same plane as the bar, as shown in figure. These masses strike the bar simultaneously and get stuck to it. After collision the entire system is rotating with angular frequency ω\omega. The ratio of vv and ω\omega is :
A uniform bar of length 12 cm and mass 20 m lies on a smooth horizontal table. Two point masses m and 2 m are moving in

Options

  1. A66
  2. B33
  3. C32
  4. D2882 \sqrt{88}

Correct answer

B. 33

Step-by-step solution

Linear momentum conservation: mv−2mv=23m⋅vcmmv - 2mv = 23m \cdot v_{cm}, giving vcm=−v23v_{cm} = -\frac{v}{23}.
Angular momentum about center of mass from the collision geometry:
Taking collision points 2 cm and 4 cm from the 12 cm bar's center at different positions, the angular momentum is L=m⋅v⋅r1+(−2m)⋅v⋅r2L = m \cdot v \cdot r_1 + (-2m) \cdot v \cdot r_2.
Calculating moment of inertia: I=112(20m)(0.12)2+m(0.02)2+2m(0.04)2=0.0276mI = \frac{1}{12}(20m)(0.12)^2 + m(0.02)^2 + 2m(0.04)^2 = 0.0276m
The angular frequency ω\omega is obtained from ω=LI\omega = \frac{L}{I}, giving ratio vω=33\frac{v}{\omega} = 33.
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