OA Explanation:

This question is based on the concept of Relative velocity.
(Refer to the figure below)

The speeds are given in Kmph, so the distance must be taken in Km and time in Hr
Let the length of Train A be \(D\) and speed of Train B be \(S\) kmph

For opposite deirection;
Relative Velocity = \((63 + S)\) Kmph

So, \(t = \frac{27}{3600} = \frac{(D + 0.5)}{(63 + S)}\)

In same deirection;
Relative Velocity = \((63 - S)\) Kmph
Note, since Train A is chasing Train B here, speed of Train A has to be greater than speed of Train B

So, \(t = \frac{162}{3600} = \frac{(D + 0.5)}{(63 - S)}\)

Equate the distance;
\(\frac{27}{3600}(63 + S) = \frac{162}{3600}(63 - S)\)
\(27(63 + S) = 162(63 - S)\)
\((63 + S) = 6(63 - S)\)
\(7S = 5(63)\)
\(S = 45\) Kmph

Put it in any equation to get D;
\(\frac{27}{3600} = \frac{(D + 0.5)}{(63 + S)}\)
\(\frac{27}{3600} = \frac{(D + 0.5)}{(63 + 45)}\)
\(\frac{27}{3600}(108) = (D + 0.5)\)
\(0.81 = (D + 0.5)\)
\(D = 0.31\) Km i.e. 310m

Hence, option A