JEE Physics 2026 Question with Solution

Given: River width is $200 \, \text{m}$, river speed is $18 \, \text{km/h}$, and boat speed in still water is $36 \, \text{km/h}$.

Find: The minimum total time for the one-round trip and the displacement along the river bank.

For minimum time, the boat must be steered perpendicular to the bank so that its full speed contributes to crossing.

Convert speeds:

$$v_{\text{river}} = 18 \times \frac{5}{18} = 5 \, \text{m/s}$$ $$v_{\text{boat}} = 36 \times \frac{5}{18} = 10 \, \text{m/s}$$

Time to cross once:

$$t_1 = \frac{200}{10} = 20 \, \text{s}$$

Time to return:

$$t_2 = \frac{200}{10} = 20 \, \text{s}$$

Total minimum time:

$$T = t_1 + t_2 = 20 + 20 = 40 \, \text{s}$$

Drift during first crossing:

$$d_1 = v_{\text{river}} \times t_1 = 5 \times 20 = 100 \, \text{m}$$

Drift during return crossing:

$$d_2 = v_{\text{river}} \times t_2 = 5 \times 20 = 100 \, \text{m}$$

Total displacement along the bank:

$$d = d_1 + d_2 = 100 + 100 = 200 \, \text{m}$$

Therefore, the journey takes $40 \, \text{s}$ and the displacement along the river bank is $200 \, \text{m}$. The correct option is D.