Question

Two men stand a certain distance apart beside a long metal fence on a still day. One man places his ear against the fence while the other gives the fence a sharp knock with a hammer. Two sounds separated by a time interval of $0.5$ second, are heard by the first man. If the velocity of sound in air is $330m{s}^{-1}$ and in the metal $5280m{s}^{-1}$, how far apart are the men?

• A. 352 m
• B. 330 m
• C. 162 m
• D. 176 m

Answer 1

The correct option is D 176 m

Let the distance between two men be $l$ metres.

Given : Speed of sound in air $v_a=330m/s$

Speed of sound in metal $v_m=5280m/s$

Let time taken by man to hear the sound through metal be $tseconds$.

Thus time taken by man to hear the sound in air $T=t+0.5seconds$

Using $l=v\times t$

$l=5280t$ .............(1)

$l=330(t+0.5)$ ..............(2)

Equating (1) and (2), $5280t=330(t+0.5)⟹t=0.033sec$

Now from (1), $l=5280\times 0.033$

Thus $l\approx 176m$