The ratio of times taken by freely falling body to cover first metre, second metre,... is

Options

A. `sqrt1:sqrt2:sqrt3`
B. `sqrt1:sqrt2 -sqrt1:sqrt3-sqrt2`
C. `sqrt2:sqrt4:sqrt8`
D. `2:3:4`

Answer

To solve the problem of finding the ratio of times taken by a freely falling body to cover the first meter, second meter, and so on, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Motion**:
A freely falling body starts from rest and accelerates downwards under the influence of gravity (g). The initial velocity (u) is 0.

2. **Using the Second Equation of Motion**:
The second equation of motion states:
\[ s = ut + \frac{1}{2} a t^2 \]
For a freely falling body, \( u = 0 \) and \( a = g \). Thus, the equation simplifies to:
\[ s = \frac{1}{2} g t^2 \]

3. **Finding the Time for Each Meter**:
The time taken to cover a distance \( s \) can be rearranged from the equation:
\[ t = \sqrt{\frac{2s}{g}} \]
We will calculate the time taken to cover the first meter (s = 1 m), the second meter (s = 2 m), and the third meter (s = 3 m).

4. **Calculating Time for Each Segment**:
- **Time to cover the first meter (t1)**:
\[ t_1 = \sqrt{\frac{2 \times 1}{g}} = \sqrt{\frac{2}{g}} \]
- **Time to cover the second meter (t2)**:
To find the time taken to cover the second meter, we need to find the total time to fall 2 meters and subtract the time to fall the first meter:
\[ t_{total\_2} = \sqrt{\frac{2 \times 2}{g}} = \sqrt{\frac{4}{g}} = 2\sqrt{\frac{1}{g}} \]
Therefore, the time to cover the second meter (t2) is:
\[ t_2 = t_{total\_2} - t_1 = 2\sqrt{\frac{1}{g}} - \sqrt{\frac{2}{g}} = \left(2 - \sqrt{2}\right)\sqrt{\frac{1}{g}} \]

- **Time to cover the third meter (t3)**:
Similarly, for the third meter:
\[ t_{total\_3} = \sqrt{\frac{2 \times 3}{g}} = \sqrt{\frac{6}{g}} = \sqrt{6}\sqrt{\frac{1}{g}} \]
The time to cover the third meter (t3) is:
\[ t_3 = t_{total\_3} - t_{total\_2} = \sqrt{6}\sqrt{\frac{1}{g}} - 2\sqrt{\frac{1}{g}} = \left(\sqrt{6} - 2\right)\sqrt{\frac{1}{g}} \]

5. **Finding the Ratios**:
Now we can find the ratios of the times taken to cover the first, second, and third meters:
\[ \text{Ratio} = \frac{t_1}{t_2} : \frac{t_2}{t_3} : \frac{t_3}{t_1} \]
Substituting the values we calculated:
\[ \text{Ratio} = \sqrt{\frac{2}{g}} : \left(2 - \sqrt{2}\right)\sqrt{\frac{1}{g}} : \left(\sqrt{6} - 2\right)\sqrt{\frac{1}{g}} \]

6. **Simplifying the Ratios**:
Since all terms have \(\sqrt{\frac{1}{g}}\), we can cancel this out:
\[ \text{Ratio} = \sqrt{2} : \left(2 - \sqrt{2}\right) : \left(\sqrt{6} - 2\right) \]

### Final Answer:
The ratio of times taken by a freely falling body to cover the first meter, second meter, and third meter is:
\[ \text{Ratio} = \sqrt{2} : (2 - \sqrt{2}) : (\sqrt{6} - 2) \]