JEE Main 2026 — Rotational Motion Question with Solution

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