question_answer

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