Solution:

To understand the ratio of the distances traveled by a freely falling body in the 1st, 2nd, 3rd, and 4th seconds, we need to consider the motion of the body under free fall.

1. Understanding Free Fall:
When an object is in free fall, it means it is only under the influence of gravity, and no other forces are acting on it. In such a scenario, the object accelerates downwards at a constant rate, known as the acceleration due to gravity (g).

2. Distance Traveled in Free Fall:
The distance traveled by a freely falling body during each second can be calculated using the equation for distance traveled with constant acceleration:

d = ut + 0.5at^2

where d is the distance traveled, u is the initial velocity (which is zero in this case), a is the acceleration (equal to g), and t is the time.

3. Analyzing the Options:
Let's calculate the distances traveled by the body in the 1st, 2nd, 3rd, and 4th seconds using the given options and see which one matches the pattern.

a) 1 : 2 : 3 : 4
- Distance in 1st second: d1 = 0.5 × g × (1^2) = 0.5g
- Distance in 2nd second: d2 = 0.5 × g × (2^2) = 2g
- Distance in 3rd second: d3 = 0.5 × g × (3^2) = 4.5g
- Distance in 4th second: d4 = 0.5 × g × (4^2) = 8g

b) 1 : 4 : 9 : 16
- Distance in 1st second: d1 = 0.5 × g × (1^2) = 0.5g
- Distance in 2nd second: d2 = 0.5 × g × (2^2) = 2g
- Distance in 3rd second: d3 = 0.5 × g × (3^2) = 4.5g
- Distance in 4th second: d4 = 0.5 × g × (4^2) = 8g

c) 1 : 3 : 5 : 7
- Distance in 1st second: d1 = 0.5 × g × (1^2) = 0.5g
- Distance in 2nd second: d2 = 0.5 × g × (2^2) = 2g
- Distance in 3rd second: d3 = 0.5 × g × (3^2) = 4.5g
- Distance in 4th second: d4 = 0.5 × g × (4^2) = 8g

d) 1 : 1 : 1 : 1
- Distance in 1st second: d1 = 0.5 × g × (1^2) = 0.5g
- Distance in 2nd second: d2 = 0.5 × g × (2^2) = 2g
- Distance in