From the top of a tower, a particle is thrown vertically downwards with a velocity of `10 m//s`. The ratio of the distances, covered by it in the `3rd` and `2nd` seconds of the motion is `("Take" g = 10 m//s^2)`.

Options

A. `5 : 7`
B. `7 : 5`
C. `3 : 6`
D. `6 : 3`

Explanation

To solve the problem, we need to find the ratio of the distances covered by a particle thrown vertically downwards in the 3rd and 2nd seconds of its motion. The initial velocity \( u \) is \( 10 \, \text{m/s} \) and the acceleration due to gravity \( g \) is \( 10 \, \text{m/s}^2 \).

### Step-by-Step Solution:

1. **Understanding the Formula**:
The distance covered by the particle in the \( n \)-th second is given by the formula:
\[ S_n = u + \frac{1}{2} g (2n - 1) \]
where \( S_n \) is the distance covered in the \( n \)-th second, \( u \) is the initial velocity, \( g \) is the acceleration due to gravity, and \( n \) is the second in which we want to find the distance.

2. **Calculate Distance in the 3rd Second (\( S_3 \))**:
For \( n = 3 \):
\[ S_3 = u + \frac{1}{2} g (2 \cdot 3 - 1) \]
Substituting \( u = 10 \, \text{m/s} \) and \( g = 10 \, \text{m/s}^2 \):
\[ S_3 = 10 + \frac{1}{2} \cdot 10 \cdot (6 - 1) = 10 + \frac{1}{2} \cdot 10 \cdot 5 = 10 + 25 = 35 \, \text{m} \]

3. **Calculate Distance in the 2nd Second (\( S_2 \))**:
For \( n = 2 \):
\[ S_2 = u + \frac{1}{2} g (2 \cdot 2 - 1) \]
Substituting the same values:
\[ S_2 = 10 + \frac{1}{2} \cdot 10 \cdot (4 - 1) = 10 + \frac{1}{2} \cdot 10 \cdot 3 = 10 + 15 = 25 \, \text{m} \]

4. **Finding the Ratio \( \frac{S_3}{S_2} \)**:
Now, we can find the ratio of the distances covered in the 3rd and 2nd seconds:
\[ \frac{S_3}{S_2} = \frac{35}{25} = \frac{7}{5} \]

5. **Final Answer**:
The ratio of the distances covered by the particle in the 3rd and 2nd seconds is:
\[ \frac{S_3}{S_2} = \frac{7}{5} \]