Q69 / 75 · JEE Main 2026 (21 January Shift 2) · Physics
Two cars $A$ and $B$ each of mass $10^{3} \mathrm{~kg}$ are moving on parallel tracks separated by a distance of 10 m, in same direction with speeds $72 \mathrm{~km} / \mathrm{h}$ and $36 \mathrm{~km} / \mathrm{h}$. The magnitude of angular momentum of $\operatorname{car} A$ with respect to $\operatorname{car} B$ is $\_\_\_\_$ J.s.
- A$2 \times 10^{5}$
- Correct answerB$10^{5}$
- C$3.6 \times 10^{5}$
- D$3 \times 10^{5}$
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).
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).
Hence, the correct answer is option: 'B'.