A train of length 442 metres crosses an electric pole in 13 seconds and crosses another train of the same length travelling in opposite direction in 17 seconds. What is the speed of the second train?

The first train has length $$L = 442\,\text{m}$$ and takes $$t_1 = 13\,\text{s}$$ to pass a stationary pole.

Speed of the first train: $$v_1 = \frac{L}{t_1} = \frac{442}{13} = 34\,\text{m/s}.$$(1)

When two trains move in opposite directions, the distance covered while they cross each other equals the sum of their lengths. Here each train is 442 m long, so the total distance is $$2L = 884\,\text{m}.$$(2)

The two trains cross in $$t_2 = 17\,\text{s},$$ so their relative speed is $$v_\text{rel} = \frac{884}{17} = 52\,\text{m/s}.$$(3)

For opposite directions, $$v_\text{rel} = v_1 + v_2.$$(4)
Using (1) and (3) in (4): $$v_2 = v_\text{rel} - v_1 = 52 - 34 = 18\,\text{m/s}.$$

Therefore, the speed of the second train is $$18\,$$\text{m/s}$$.$

Option B which is: 18 m/s