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`

Answer

To solve the problem, we need to calculate the distances covered by the particle in the 2nd and 3rd seconds of its motion and then find the ratio of these distances.

### Step 1: Understanding the motion
The particle is thrown downwards with an initial velocity \( u = 10 \, \text{m/s} \) and it experiences a constant acceleration due to gravity \( g = 10 \, \text{m/s}^2 \).

### Step 2: Formula for distance covered in nth second
The distance covered in the nth second can be calculated using the formula:
\[ s_n = u + \frac{1}{2} a (2n - 1) \]
Here, \( a = g \) (acceleration due to gravity).

### Step 3: Calculate distance covered in the 2nd second
For \( n = 2 \):
\[ s_2 = u + \frac{1}{2} g (2 \cdot 2 - 1) \]
Substituting the values:
\[ s_2 = 10 + \frac{1}{2} \cdot 10 \cdot (4 - 1) \]
\[ s_2 = 10 + \frac{1}{2} \cdot 10 \cdot 3 \]
\[ s_2 = 10 + 15 = 25 \, \text{m} \]

### Step 4: Calculate distance covered in the 3rd second
For \( n = 3 \):
\[ s_3 = u + \frac{1}{2} g (2 \cdot 3 - 1) \]
Substituting the values:
\[ s_3 = 10 + \frac{1}{2} \cdot 10 \cdot (6 - 1) \]
\[ s_3 = 10 + \frac{1}{2} \cdot 10 \cdot 5 \]
\[ s_3 = 10 + 25 = 35 \, \text{m} \]

### Step 5: Calculate the ratio of distances
Now, we find the ratio of the distances covered in the 3rd and 2nd seconds:
\[ \text{Ratio} = \frac{s_3}{s_2} = \frac{35}{25} = \frac{7}{5} \]

### Final Answer
The ratio of the distances covered by the particle in the 3rd and 2nd seconds of motion is \( \frac{7}{5} \).

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