The given problem involves water droplets falling from a tap located at a height of 27 meters above the ground. We are asked to calculate the position of the 2nd and 3rd droplets when the 4th droplet is about to fall.

Let's analyze the problem step by step:

1. Finding the time interval between each droplet:
Since the water droplets fall from the tap at regular time intervals, we need to find the time it takes for one droplet to fall from the tap to the ground.

Using the equation of motion, h = (1/2)gt^2, where h is the height, g is the acceleration due to gravity (approximately 9.8 m/s^2), and t is the time, we can solve for t:

27 = (1/2)(9.8)t^2
54 = 9.8t^2
t^2 = 54/9.8
t = √(54/9.8)
t ≈ 2.36 seconds

Therefore, each water droplet takes approximately 2.36 seconds to fall from the tap to the ground.

2. Calculating the position of the 2nd droplet:
Given that the 1st droplet just touches the ground, the 2nd droplet would have fallen for a time period of 2.36 seconds.

Using the equation of motion, h = (1/2)gt^2, we can find the distance traveled by the 2nd droplet in 2.36 seconds:

h = (1/2)(9.8)(2.36)^2
h ≈ 27.1 meters

Therefore, the position of the 2nd droplet is approximately 27.1 meters below the tap.

3. Calculating the position of the 3rd droplet:
Since the 3rd droplet falls after the 2nd droplet, it would have fallen for a time period of 2.36 seconds as well.

Using the equation of motion, we can find the distance traveled by the 3rd droplet:

h = (1/2)(9.8)(2.36)^2
h ≈ 27.1 meters

Therefore, the position of the 3rd droplet is also approximately 27.1 meters below the tap.

In conclusion, the position of the 2nd and 3rd droplets is approximately 27.1 meters below the tap.