A ball is released from the top of a tower. The ratio of work done by force of gravity in 1st second, 2nd second and 3rd second of the motion of ball is

Options

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

Answer

To find the ratio of work done by the force of gravity on a ball released from the top of a tower during the first, second, and third seconds of its motion, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Motion**:
The ball is released from rest, so its initial velocity (u) is 0. The only force acting on it is gravity, which provides a constant acceleration (g).

2. **Distance Covered in Each Second**:
We need to calculate the distance covered by the ball in the first, second, and third seconds. The formula for the distance covered in the nth second is given by:
\[ s_n = u + \frac{g}{2}(2n - 1) \]
Since \(u = 0\), the formula simplifies to:
\[ s_n = \frac{g}{2}(2n - 1) \]

3. **Calculating Distances**:
- For the **1st second** (n = 1):
\[ s_1 = \frac{g}{2}(2 \cdot 1 - 1) = \frac{g}{2}(1) = \frac{g}{2} \]
- For the **2nd second** (n = 2):
\[ s_2 = \frac{g}{2}(2 \cdot 2 - 1) = \frac{g}{2}(3) = \frac{3g}{2} \]
- For the **3rd second** (n = 3):
\[ s_3 = \frac{g}{2}(2 \cdot 3 - 1) = \frac{g}{2}(5) = \frac{5g}{2} \]

4. **Finding the Ratio of Distances**:
The distances covered in the first, second, and third seconds are:
- \(s_1 = \frac{g}{2}\)
- \(s_2 = \frac{3g}{2}\)
- \(s_3 = \frac{5g}{2}\)

Therefore, the ratio of distances \(s_1 : s_2 : s_3\) is:
\[ \frac{g/2}{3g/2} : \frac{3g/2}{5g/2} = 1 : 3 : 5 \]

5. **Work Done by Gravity**:
The work done by the force of gravity (W) is given by:
\[ W = F \cdot s = mg \cdot s \]
Since the force of gravity (mg) is constant, the work done is directly proportional to the distance covered. Hence, the ratio of work done in the first, second, and third seconds is the same as the ratio of distances:
\[ W_1 : W_2 : W_3 = s_1 : s_2 : s_3 = 1 : 3 : 5 \]

### Final Answer:
The ratio of work done by the force of gravity in the first, second, and third seconds of the motion of the ball is \(1 : 3 : 5\).

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