A stone is dropped in a well which is 19.6 m deep. Echo sound is heard after 2.06 sec (after dropping) then the velocity of sound is

Options

A. 332.6 m/sec
B. 326.7 m/sec
C. 300.4 m/sec
D. 290.5 m/sec

Answer

To solve the problem step by step, we will break down the process of calculating the velocity of sound based on the information provided.

### Step 1: Understand the Problem
A stone is dropped into a well that is 19.6 m deep. The total time taken from when the stone is dropped until the echo is heard is 2.06 seconds. We need to find the velocity of sound in air.

### Step 2: Define Variables
- Depth of the well (h) = 19.6 m
- Total time (T) = 2.06 s
- Time taken for the stone to fall (T1)
- Time taken for the sound to travel back up (T2)
- Velocity of sound (v)

### Step 3: Calculate Time for the Stone to Fall (T1)
The stone is dropped from rest, so we can use the equation of motion:
\[ h = \frac{1}{2} g T_1^2 \]
Where:
- g = 9.8 m/s² (acceleration due to gravity)

Rearranging the equation to solve for \( T_1 \):
\[ T_1^2 = \frac{2h}{g} \]
\[ T_1 = \sqrt{\frac{2h}{g}} \]

Substituting the values:
\[ T_1 = \sqrt{\frac{2 \times 19.6}{9.8}} \]
\[ T_1 = \sqrt{\frac{39.2}{9.8}} \]
\[ T_1 = \sqrt{4} \]
\[ T_1 = 2 \text{ seconds} \]

### Step 4: Calculate Time for Sound to Travel Up (T2)
The total time for the stone to fall and the sound to travel back up is given by:
\[ T = T_1 + T_2 \]
Thus,
\[ T_2 = T - T_1 \]
Substituting the known values:
\[ T_2 = 2.06 - 2 \]
\[ T_2 = 0.06 \text{ seconds} \]

### Step 5: Calculate the Velocity of Sound (v)
The sound travels the same distance (19.6 m) back up to the top of the well. We can use the formula:
\[ v = \frac{h}{T_2} \]
Substituting the values:
\[ v = \frac{19.6}{0.06} \]
\[ v = 326.67 \text{ m/s} \]

### Final Answer
The velocity of sound is approximately **326.67 m/s**.

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