Question

The ratio of times taken by a freely falling body to travel first 5m, second 5m, third 5m distances is

• A. $1 : \sqrt{2} : \sqrt{3}$
• B. $1 : \sqrt{2}-1 : \sqrt{3}-2$
• C. $1 : \sqrt{3} : \sqrt{5}$
• D. $1 : \sqrt{2}-1 : \sqrt{3}-\sqrt{2}$

Hint:

Galileo's Law of Odd Numbers states that for equal *time intervals*, distances are in ratio $1:3:5$. Conversely, for equal *distance intervals*, the time intervals are in ratio $1 : (\sqrt{2}-1) : (\sqrt{3}-\sqrt{2}) : \dots$

Answer 1

The Correct Option is D

Step 1: Key Formula:
For a freely falling body starting from rest ($u=0$), the distance covered is given by $s = \frac{1}{2}gt^2$. Therefore, time taken to cover distance $s$ is $t = \sqrt{\frac{2s}{g}} \propto \sqrt{s}$.
Step 2: Calculate Cumulative Times:
Let $d = 5\text{ m}$. Time to travel first $d$ (Total distance $d$): $t_1 \propto \sqrt{d} \propto \sqrt{1}$ Time to travel first $2d$ (Total distance $10\text{ m}$): $t_2 \propto \sqrt{2d} \propto \sqrt{2}$ Time to travel first $3d$ (Total distance $15\text{ m}$): $t_3 \propto \sqrt{3d} \propto \sqrt{3}$
Step 3: Calculate Interval Times:
Time for 1st 5m ($\Delta t_1$) = $t_1 \propto \sqrt{1}$ Time for 2nd 5m ($\Delta t_2$) = $t_2 - t_1 \propto \sqrt{2} - \sqrt{1}$ Time for 3rd 5m ($\Delta t_3$) = $t_3 - t_2 \propto \sqrt{3} - \sqrt{2}$
Step 4: Ratio:
The ratio is $1 : (\sqrt{2}-1) : (\sqrt{3}-\sqrt{2})$.