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 3.5s respectively, the distance between the cliffs is (Velocity of sound in air = 340 ms?1) [EAMCET (Med.) 2000]

A)            1190 m                                    

B)            850 m

C)            595 m                                      

D)            510 m

Correct Answer: B

Solution :

            \[2{{d}_{1}}+2{{d}_{2}}=v\times {{t}_{1}}+v\times {{t}_{2}}\]Þ \[2({{d}_{1}}+{{d}_{2}})=v({{t}_{1}}+{{t}_{2}})\]                    \[{{d}_{1}}+{{d}_{2}}=\frac{v({{t}_{1}}+{{t}_{2}})}{2}\]=\[\frac{340\times (1.5+3.5)}{2}=850\] m.