Question

A body of mass 0.2 kg is falling freely from a height of 180 m from the ground. The ratio of the works done by the gravitational force in the first two seconds and in the next two seconds of motion of the body is

• A. 1 : 3
• B. 1 : 2
• C. 2 : 3
• D. 3 : 4

Hint:

{Galileo's Law of Odd Numbers:} For a body falling from rest, distances travelled in equal successive time intervals are in the ratio $1 : 3 : 5 : 7 \dots$ Here, 1st interval is $0-2s$ (Ratio part 1) and 2nd interval is $2-4s$ (Ratio part 3). The answer is directly 1:3.

Answer 1

The Correct Option is A

Step 1: Relationship between Work and Distance:
Work done by gravity is given by $W = mgh$, where $h$ is the vertical distance fallen. Since $m$ and $g$ are constant, the ratio of work done is equal to the ratio of distances fallen in the given time intervals ($W \propto h$).
Step 2: Calculate Distances Fallen:
Using the equation of motion for free fall ($u=0$): $h = \frac{1}{2}gt^2$.
Distance in first 2 seconds ($t=0$ to $t=2$): \[ h_1 = \frac{1}{2} g (2)^2 = 2g \]
Distance in first 4 seconds ($t=0$ to $t=4$): \[ h_{total} = \frac{1}{2} g (4)^2 = 8g \]
Distance in "next two seconds" ($t=2$ to $t=4$): \[ h_2 = h_{total} - h_1 = 8g - 2g = 6g \]

Step 3: Calculate the Ratio:
Ratio of Work = Ratio of Distances ($h_1 : h_2$). \[ \text{Ratio} = \frac{h_1}{h_2} = \frac{2g}{6g} = \frac{1}{3} \]
Step 4: Final Answer:
The ratio is 1:3.