An iron block is dropped into a deep well. Sound of splash is heard after 4.23 s. If the depth of the well is 78. 4 m , then find the speed of sound in air `(g=9.8 m//s^(2))`

Options

A. 300 m/s
B. 320 m/s
C. 280 m/s
D. 340.8m/s

Answer

To solve the problem step by step, we will break down the process of finding the speed of sound in air after an iron block is dropped into a well.

### Step 1: Understand the Problem
We have an iron block dropped into a well, and we need to find the speed of sound in air given that the total time taken for the sound of the splash to be heard is 4.23 seconds, and the depth of the well is 78.4 meters.

### Step 2: Define the Variables
Let:
- \( t_1 \) = time taken for the block to fall to the water level
- \( t_2 \) = time taken for the sound to travel back up to the top of the well
- Total time = \( t_1 + t_2 = 4.23 \) seconds
- Depth of the well = \( s = 78.4 \) meters
- Acceleration due to gravity = \( g = 9.8 \, \text{m/s}^2 \)

### Step 3: Calculate \( t_1 \)
Using the equation of motion for the falling block:
\[ s = ut + \frac{1}{2} a t^2 \]
Since the initial velocity \( u = 0 \) (the block is dropped), the equation simplifies to:
\[ s = \frac{1}{2} g t_1^2 \]
Substituting the known values:
\[ 78.4 = \frac{1}{2} \times 9.8 \times t_1^2 \]
\[ 78.4 = 4.9 t_1^2 \]
Now, solving for \( t_1^2 \):
\[ t_1^2 = \frac{78.4}{4.9} \approx 16 \]
Taking the square root gives:
\[ t_1 = \sqrt{16} = 4 \, \text{seconds} \]

### Step 4: Calculate \( t_2 \)
Now that we have \( t_1 \), we can find \( t_2 \):
\[ t_2 = 4.23 - t_1 = 4.23 - 4 = 0.23 \, \text{seconds} \]

### Step 5: Calculate the Speed of Sound
We know the distance the sound travels (which is the depth of the well) and the time taken for that distance:
\[ \text{Speed of sound} = \frac{\text{Distance}}{\text{Time}} = \frac{78.4 \, \text{m}}{0.23 \, \text{s}} \]
Calculating this gives:
\[ \text{Speed of sound} \approx 340.87 \, \text{m/s} \]

### Conclusion
The speed of sound in air is approximately \( 340.87 \, \text{m/s} \).