Drops of water fall at regular intervals from the roof of a building of height H = 5 m, the first drop strikes the ground at the same moment when the fifth drop detaches itself from the roof. Then ratio of distances as first drop reaches the ground are :

Options

A. `1: 5:7:3`
B. `1:3:5:7`
C. `1:3:7:5`
D. `1:2:3:4`

Explanation

To solve the problem, we need to analyze the motion of the water drops falling from the roof of a building of height \( H = 5 \, \text{m} \). The first drop strikes the ground at the same moment when the fifth drop detaches itself from the roof. We will find the ratio of distances traveled by the drops when the first drop reaches the ground.

### Step-by-Step Solution:

1. **Understanding the Time Intervals**:
- Let the time interval between the detachment of each drop be \( t_0 \).
- The first drop falls for a total time of \( 4t_0 \) (since it falls while the second, third, fourth, and fifth drops are detached).
- The second drop falls for \( 3t_0 \), the third for \( 2t_0 \), the fourth for \( t_0 \), and the fifth drop has just detached and has not fallen yet.

2. **Using the Equation of Motion**:
- The distance fallen by an object under gravity is given by the equation:
\[ s = \frac{1}{2} g t^2 \]
- Here, \( g \) is the acceleration due to gravity.

3. **Calculating Distances for Each Drop**:
- For the first drop (falling for \( 4t_0 \)):
\[ s_1 = \frac{1}{2} g (4t_0)^2 = 8g t_0^2 \]
- For the second drop (falling for \( 3t_0 \)):
\[ s_2 = \frac{1}{2} g (3t_0)^2 = \frac{9}{2} g t_0^2 \]
- For the third drop (falling for \( 2t_0 \)):
\[ s_3 = \frac{1}{2} g (2t_0)^2 = 2g t_0^2 \]
- For the fourth drop (falling for \( t_0 \)):
\[ s_4 = \frac{1}{2} g (t_0)^2 = \frac{1}{2} g t_0^2 \]
- The fifth drop has not fallen yet, so:
\[ s_5 = 0 \]

4. **Finding the Distances Between Drops**:
- The distances between the drops when the first drop reaches the ground are:
- Distance between the first and second drop:
\[ d_{1,2} = s_1 - s_2 = 8g t_0^2 - \frac{9}{2} g t_0^2 = \frac{16g t_0^2 - 9g t_0^2}{2} = \frac{7g t_0^2}{2} \]
- Distance between the second and third drop:
\[ d_{2,3} = s_2 - s_3 = \frac{9}{2} g t_0^2 - 2g t_0^2 = \frac{9g t_0^2 - 4g t_0^2}{2} = \frac{5g t_0^2}{2} \]
- Distance between the third and fourth drop:
\[ d_{3,4} = s_3 - s_4 = 2g t_0^2 - \frac{1}{2} g t_0^2 = \frac{4g t_0^2 - g t_0^2}{2} = \frac{3g t_0^2}{2} \]
- Distance between the fourth and fifth drop:
\[ d_{4,5} = s_4 - s_5 = \frac{1}{2} g t_0^2 - 0 = \frac{1}{2} g t_0^2 \]

5. **Calculating the Ratios**:
- The distances between the drops are:
- \( d_{1,2} = \frac{7g t_0^2}{2} \)
- \( d_{2,3} = \frac{5g t_0^2}{2} \)
- \( d_{3,4} = \frac{3g t_0^2}{2} \)
- \( d_{4,5} = \frac{1}{2} g t_0^2 \)

- The ratio of distances is:
\[ 7 : 5 : 3 : 1 \]

### Final Answer:
The ratio of distances as the first drop reaches the ground is:
\[ \text{Ratio} = 7 : 5 : 3 : 1 \]