Wave Mechanics Question 830
Question: A man is standing between two parallel cliffs and fires a gun. If he hears first and second echoes after 1.5 s and 3.5s respectively, the distance between the cliffs is (Velocity of sound in air = 340 ms?1)
[EAMCET (Med.) 2000]
Options:
A) 1190 m
B) 850 m
C) 595 m
D) 510 m
Show Answer
Answer:
Correct Answer: B
Solution:
$ 2d _{1}+2d _{2}=v\times t _{1}+v\times t _{2} $
Therefore $ 2(d _{1}+d _{2})=v(t _{1}+t _{2}) $
$ d _{1}+d _{2}=\frac{v(t _{1}+t _{2})}{2} $ =$ \frac{340\times (1.5+3.5)}{2}=850 $ m.
BETA
