Read the given passage and select the correct option for the following questions. The sound returning back towards the source after suffering reflection from a distant obstacle like wall, hill etc. is called an echo. A cracker is fired by a child. The child hears an echo from a cliff 4 seconds after the sound produced by a cracker. The speed of sound in air at `20^(@)C` is 344 `ms^(-1)`.
How far is the cliff from the child?

Options

A. 344 m
B. 688 m
C. 200 m
D. 300 m

Explanation

To find the distance between the child and the cliff, we can follow these steps:

### Step 1: Understand the Problem
The child hears an echo 4 seconds after firing a cracker. The speed of sound in air at 20°C is given as 344 m/s. We need to find the distance to the cliff.

### Step 2: Use the Concept of Echo
When the child fires the cracker, the sound travels to the cliff and reflects back. The total time taken for the sound to travel to the cliff and back is 4 seconds.

### Step 3: Calculate the Total Distance
Since the sound travels to the cliff and back, the total distance traveled by the sound is twice the distance to the cliff (D). Therefore, we can express this as:
\[ \text{Total Distance} = 2D \]

### Step 4: Relate Distance, Speed, and Time
We know that:
\[ \text{Distance} = \text{Speed} \times \text{Time} \]
In this case, the total distance traveled by the sound is:
\[ 2D = \text{Speed of Sound} \times \text{Total Time} \]
Substituting the known values:
\[ 2D = 344 \, \text{m/s} \times 4 \, \text{s} \]

### Step 5: Calculate the Total Distance
Now, calculate the total distance:
\[ 2D = 344 \times 4 = 1376 \, \text{m} \]

### Step 6: Solve for D
To find the distance to the cliff (D), we divide the total distance by 2:
\[ D = \frac{1376}{2} = 688 \, \text{m} \]

### Conclusion
The distance from the child to the cliff is **688 meters**.

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