JEE Physics 2026 Question with Solution

Given: Two cars $A$ and $B$ each have mass $10^3 \, \text{kg}$. They move on parallel tracks separated by $10 \, \text{m}$ with speeds $72 \, \text{km/h}$ and $36 \, \text{km/h}$ in the same direction.

Find: The magnitude of angular momentum of car $A$ with respect to car $B$.

For relative motion along parallel lines, the angular momentum magnitude is

$$L = m \, v_{rel} \, d$$

where $v_{rel}$ is the relative speed and $d$ is the perpendicular distance between the tracks.

Convert the speeds into SI units:

$$v_A = 72 \times \frac{5}{18} = 20 \, \text{m/s}$$ $$v_B = 36 \times \frac{5}{18} = 10 \, \text{m/s}$$

So, the relative speed of $A$ with respect to $B$ is

$$v_{rel} = v_A - v_B = 20 - 10 = 10 \, \text{m/s}$$

Now substitute the values:

$$L = 10^3 \times 10 \times 10 = 10^5 \; \text{kg}\cdot\text{m}^2/\text{s}$$

This is numerically equal to $10^5 \; \text{J.s}$.

Therefore, the angular momentum is $10^5 \; \text{J.s}$, so the correct option is B.

Given: The two cars move in the same direction on parallel tracks.

Find: Why relative speed is used here.

From the frame of car $B$, car $A$ moves with speed $v_A - v_B$ along a straight line parallel to the original track. The perpendicular distance of this line of motion from $B$ remains $10 \, \text{m}$.

Hence the angular momentum magnitude about $B$ is the product of mass, relative speed, and perpendicular distance:

$$L = m \, (v_A - v_B) \, d$$

Using

$$m = 10^3 \, \text{kg}, \quad v_A - v_B = 10 \, \text{m/s}, \quad d = 10 \, \text{m}$$

we get

$$L = 10^5 \; \text{J.s}$$

So the correct option is B.