question_answer

There are two walls A and B which are 180 m apart from each other. A boy standing at 60 m away from wall A claps his hands once. If the speed of sound in air is \[330\text{ }m\text{ }{{s}^{-1}}\], what is the time interval between the first and the second echo that he hears?

A) 0.36 s              

B) 0.24 s  

C)        0.4 s    

D)        1 s                   

Correct Answer: A

Solution :

Let \[{{t}_{1}}\] be the time taken between the clap and the first echo from wall A. Let \[{{t}_{2}}\] be the time taken between the clap and the second echo from wall B. \[{{t}_{1}}=\frac{(2\times 60)}{330}s;{{t}_{2}}=\frac{(2\times 120)}{330}s\] The time interval \[={{t}_{2}}-{{t}_{1}}=\frac{(2\times 120)}{330}-\frac{(2\times 60)}{330}=0.36s\]