question_answer

\[A\] and \[B\] are two hills at a distance 510 m apart. A person standing between the hills claps his hands and hears two echoes at the end of 1 s and 2 s respectively. The velocity of sound in air is:                [BHU PMT-2004]

A)  225 m/s                              

B)  340 m/s

C)  510 m/s                              

D)  1020 m/s

Correct Answer: B

Solution :

                 Key Idea: Time taken by echo to reach the person is directly proportional to distance. Let distance of person from hill A be x meter, then from hill B is (510-x)m. \[\therefore \]  \[\frac{x}{510-x}=\frac{{{\left( time \right)}_{1}}}{{{\left( time \right)}_{2}}}=\frac{1}{2}\] \[\Rightarrow \]               \[x=170\,\,m\] Time taken for the echo to travel 170 m is given by \[\frac{1}{2}=0.5\,s.\] \[\therefore \]  \[\text{Velocity}=\frac{\text{distance}}{\text{time}}=\frac{170}{0.5}=340\,\,m/s\]