A body falls from a height `h = 200 m` (at New Delhi). The ratio of distance travelled in each `2 sec` during `t = 0` to `t = 6` seconds of the journey is.

Options

A. `1 : 4 : 9`
B. `1 : 2 : 4`
C. `1 : 3 : 5`
D. `1 : 2 : 3`

Explanation

To solve the problem, we need to calculate the distance traveled by a body falling freely under gravity in specific time intervals and then find the ratio of these distances.

### Step-by-Step Solution:

1. **Understand the motion of the body**:
The body is falling freely from a height of 200 m. The initial velocity (u) is 0 m/s, and the acceleration due to gravity (g) is approximately 9.81 m/s².

2. **Use the formula for distance traveled**:
The distance traveled by an object under uniform acceleration is given by the formula:
\[ s = ut + \frac{1}{2} g t^2 \]
Since the initial velocity (u) is 0, the formula simplifies to:
\[ s = \frac{1}{2} g t^2 \]

3. **Calculate distances for each time interval**:
- **For \( t = 0 \) seconds**:
\[ s(0) = \frac{1}{2} g (0)^2 = 0 \, \text{m} \]
- **For \( t = 2 \) seconds**:
\[ s(2) = \frac{1}{2} g (2)^2 = \frac{1}{2} \times 9.81 \times 4 = 19.62 \, \text{m} \]
- **For \( t = 4 \) seconds**:
\[ s(4) = \frac{1}{2} g (4)^2 = \frac{1}{2} \times 9.81 \times 16 = 78.48 \, \text{m} \]
- **For \( t = 6 \) seconds**:
\[ s(6) = \frac{1}{2} g (6)^2 = \frac{1}{2} \times 9.81 \times 36 = 176.58 \, \text{m} \]

4. **Calculate distances traveled in each 2-second interval**:
- **Distance from \( t = 0 \) to \( t = 2 \)**:
\[ d_1 = s(2) - s(0) = 19.62 - 0 = 19.62 \, \text{m} \]
- **Distance from \( t = 2 \) to \( t = 4 \)**:
\[ d_2 = s(4) - s(2) = 78.48 - 19.62 = 58.86 \, \text{m} \]
- **Distance from \( t = 4 \) to \( t = 6 \)**:
\[ d_3 = s(6) - s(4) = 176.58 - 78.48 = 98.1 \, \text{m} \]

5. **Find the ratio of distances**:
We need to find the ratio \( d_1 : d_2 : d_3 \):
\[ d_1 : d_2 : d_3 = 19.62 : 58.86 : 98.1 \]

To simplify this ratio, we can divide each term by the smallest distance (19.62):
\[ \frac{19.62}{19.62} : \frac{58.86}{19.62} : \frac{98.1}{19.62} \approx 1 : 3 : 5 \]

### Final Answer:
The ratio of distances traveled in each 2 seconds during \( t = 0 \) to \( t = 6 \) seconds is:
\[ \boxed{1 : 3 : 5} \]