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JEE Main 202621 JanuaryEvening Shift
Exam
JEE Main
Year
2026
Date
21 January
Shift
Evening Shift
Subject
Physics
Chapter
Rotational Motion

JEE Main 2026 · 21 January · Evening Shift · Physics · Rotational Motion

Two cars A and B each of mass 10^ 3 ~kg are moving on parallel...

Two cars AA and BB each of mass 103 kg10^{3} \mathrm{~kg} are moving on parallel tracks separated by a distance of 10 m, in same direction with speeds 72 km/h72 \mathrm{~km} / \mathrm{h} and 36 km/h36 \mathrm{~km} / \mathrm{h}. The magnitude of angular momentum of car⁡A\operatorname{car} A with respect to car⁡B\operatorname{car} B is ____\_\_\_\_ J.s.

Options

  1. A2×1052 \times 10^{5}
  2. B10510^{5}
  3. C3.6×1053.6 \times 10^{5}
  4. D3×1053 \times 10^{5}

Correct answer

B. 10510^{5}

Step-by-step solution

Mass of each car m=103m = 10^3 kg.
Speed of car AA, vA=72v_A = 72 km/h =72×518=20= 72 \times \frac{5}{18} = 20 m/s.
Speed of car BB, vB=36v_B = 36 km/h =36×518=10= 36 \times \frac{5}{18} = 10 m/s.
Relative velocity of car AA with respect to car BB is vAB=vA−vB=20−10=10v_{AB} = v_A - v_B = 20 - 10 = 10 m/s.
The cars are moving on parallel tracks separated by a perpendicular distance d=10d = 10 m.
The magnitude of angular momentum of car AA with respect to car BB is given by L=m×vAB×dL = m \times v_{AB} \times d.
Substituting the values: L=103×10×10=105L = 10^3 \times 10 \times 10 = 10^5 J.s.
Thus, the correct option is (2).
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