The ratio of the distance through which a body falls in 4th, 5th and 6th second is starting from rest

Options

A. `7 : 9 : 11`
B. `4 : 5 : 6`
C. `5 : 7 :9`
D. `6: 8:10`

Explanation

To find the ratio of the distances through which a body falls in the 4th, 5th, and 6th seconds, we can use the formula for the distance traveled in the t-th second when starting from rest.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- We need to find the distance fallen in the 4th, 5th, and 6th seconds of free fall. The object starts from rest, so the initial velocity (u) is 0.

2. **Formula for Distance in the t-th Second**:
- The distance fallen in the t-th second can be calculated using the formula:
\[ s_t = \frac{g}{2} (2t - 1) \]
where \( g \) is the acceleration due to gravity.

3. **Calculate the Distance for Each Second**:
- **For the 4th second (t = 4)**:
\[ s_4 = \frac{g}{2} (2 \cdot 4 - 1) = \frac{g}{2} (8 - 1) = \frac{g}{2} \cdot 7 = \frac{7g}{2} \]

- **For the 5th second (t = 5)**:
\[ s_5 = \frac{g}{2} (2 \cdot 5 - 1) = \frac{g}{2} (10 - 1) = \frac{g}{2} \cdot 9 = \frac{9g}{2} \]

- **For the 6th second (t = 6)**:
\[ s_6 = \frac{g}{2} (2 \cdot 6 - 1) = \frac{g}{2} (12 - 1) = \frac{g}{2} \cdot 11 = \frac{11g}{2} \]

4. **Finding the Ratio**:
- Now, we can write the ratio of the distances fallen in the 4th, 5th, and 6th seconds:
\[ s_4 : s_5 : s_6 = \frac{7g}{2} : \frac{9g}{2} : \frac{11g}{2} \]
- Since \( \frac{g}{2} \) is common in all terms, we can simplify the ratio:
\[ 7 : 9 : 11 \]

### Final Answer:
The ratio of the distances through which the body falls in the 4th, 5th, and 6th seconds is:
\[ \boxed{7 : 9 : 11} \]