A splash in heard `3.12` s after a stone is dropped into a well 45 m deep. The speed of sound in air is `[ g =10 ms ^(-2)]`

Options

A. `330 ms ^(-1)`
B. `375ms ^(-1)`
C. `340ms ^(-1)`
D. `346ms ^(-1)`

Answer

To solve the problem, we need to analyze the motion of the stone dropped into the well and the time taken for the sound of the splash to travel back up to the top of the well.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- A stone is dropped into a well that is 45 m deep.
- The total time taken from when the stone is dropped to when the splash is heard is 3.12 seconds.
- We need to find the speed of sound in air.

2. **Setting Up the Equations**:
- Let \( t_1 \) be the time taken for the stone to fall to the water surface.
- Let \( t_2 \) be the time taken for the sound to travel back up to the top of the well.
- According to the problem, we have:
\[ t_1 + t_2 = 3.12 \text{ seconds} \]

3. **Calculating the Time for the Stone to Fall**:
- The stone is dropped, so its initial velocity \( u = 0 \).
- The distance fallen by the stone is 45 m.
- Using the equation of motion:
\[ s = ut + \frac{1}{2}gt^2 \]
where \( s = 45 \) m, \( g = 10 \, \text{m/s}^2 \), we can substitute:
\[ 45 = 0 \cdot t_1 + \frac{1}{2} \cdot 10 \cdot t_1^2 \]
Simplifying this gives:
\[ 45 = 5t_1^2 \implies t_1^2 = \frac{45}{5} = 9 \implies t_1 = 3 \text{ seconds} \]

4. **Finding the Time for Sound to Travel**:
- Now we can find \( t_2 \):
\[ t_2 = 3.12 - t_1 = 3.12 - 3 = 0.12 \text{ seconds} \]

5. **Calculating the Speed of Sound**:
- The distance the sound travels is the same depth of the well, which is 45 m.
- The speed of sound \( v \) can be calculated using the formula:
\[ v = \frac{\text{distance}}{\text{time}} = \frac{45 \text{ m}}{t_2} \]
Substituting \( t_2 = 0.12 \) seconds:
\[ v = \frac{45}{0.12} = 375 \text{ m/s} \]

### Final Answer:
The speed of sound in air is **375 m/s**.