Given:
Distance between tap and ground = 5m
Time between two drops = constant

To find:
Height of the second drop from the ground when the first drop touches the ground.

Solution:
Let's assume that the time between two drops is 't' seconds.
Distance covered by the first drop in t seconds = 5m (distance between tap and ground)
Distance covered by the second drop in t seconds = ?
Distance covered by the third drop in t seconds = ?

Since the third drop is leaving the tap at the instant the first drop touches the ground, the time between the first and third drop is 2t seconds.

Distance covered by the third drop in 2t seconds = 5m (distance between tap and ground)
Distance covered by the second drop in 2t seconds = ?

We know that the distance covered by a freely falling body in 't' seconds is given by the formula:
d = 1/2 gt^2, where g is the acceleration due to gravity (9.8m/s^2)

Using this formula, we can find the distances covered by the drops in 't' and '2t' seconds.

Distance covered by the first drop in t seconds = 1/2 * 9.8 * t^2 = 4.9t^2
Distance covered by the third drop in 2t seconds = 1/2 * 9.8 * (2t)^2 = 19.6t^2

Since the third drop covers the distance of 5m in 2t seconds, we can equate the distance covered by the third drop to 5m.

19.6t^2 = 5
t^2 = 5/19.6
t = 0.71 seconds (approx)

Using this value of 't', we can find the distance covered by the second drop in 't' seconds.

Distance covered by the second drop in t seconds = 1/2 * 9.8 * t^2 = 4.9 * 0.71^2 = 2.5m

Therefore, the height of the second drop from the ground when the first drop touches the ground is 2.5m.

Hence, option (B) is the correct answer.