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 find the speed of sound in air, we need to break down the problem into two parts: the time it takes for the iron block to fall to the water and the time it takes for the sound to travel back up the well.

### Step 1: Understand the total time
The total time for the event is given as 4.23 seconds. This time includes:
- The time taken for the block to fall (t1)
- The time taken for the sound to travel back up (t2)

Thus, we can write:
\[ t_{\text{total}} = t_1 + t_2 \]
\[ 4.23 = t_1 + t_2 \]

### Step 2: Calculate the time taken for the block to fall (t1)
The block is dropped from rest, so we can use the equation of motion:
\[ s = ut + \frac{1}{2} g t_1^2 \]
where:
- \( s = 78.4 \, \text{m} \) (depth of the well)
- \( u = 0 \, \text{m/s} \) (initial velocity)
- \( g = 9.8 \, \text{m/s}^2 \) (acceleration due to gravity)

Substituting the values:
\[ 78.4 = 0 + \frac{1}{2} \cdot 9.8 \cdot t_1^2 \]
\[ 78.4 = 4.9 t_1^2 \]

Now, solving for \( t_1^2 \):
\[ t_1^2 = \frac{78.4}{4.9} \]
\[ t_1^2 = 16 \]
\[ t_1 = \sqrt{16} = 4 \, \text{s} \]

### Step 3: Calculate the time taken for sound to travel back up (t2)
Now we can find \( t_2 \):
\[ t_2 = t_{\text{total}} - t_1 \]
\[ t_2 = 4.23 - 4 \]
\[ t_2 = 0.23 \, \text{s} \]

### Step 4: Calculate the speed of sound in air
The speed of sound can be calculated using the formula:
\[ \text{Speed} = \frac{\text{Distance}}{\text{Time}} \]
Here, the distance is the depth of the well (78.4 m) and the time is \( t_2 \) (0.23 s):
\[ \text{Speed} = \frac{78.4}{0.23} \]

Calculating this gives:
\[ \text{Speed} \approx 340 \, \text{m/s} \]

### Final Answer
The speed of sound in air is approximately **340 m/s**.

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