A man is standing between two parallel cliffs and fires a gun. If he hears first and second echoes after 1.5 s and 3.5 s respectively the distance between the cliffs is : (Velocity of sound in air\[=340m{{s}^{-1}}\])
A) 190 m
B) 850 m
C) 591 m
D) 510 m
Correct Answer:
B
Solution :
Let the person is at point P. Time taken in first echo, \[{{t}_{1}}=1.5\,s\] \[\therefore \] \[x=\frac{V\times {{t}_{1}}}{2}\] \[=\frac{340\times 7.5}{2}=255\,m\] Similarly, \[y=\frac{V\times {{t}_{2}}}{2}\] \[=\frac{340\times 3.5}{2}\] \[=595\,m\] Distance between two cliffs \[=\text{ }x\text{ }+\text{ }y\] = 255 + 595 = 850 m