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Q. Two boys stand close to a long straight metal pipe and some distance from each other. One boy fires a gun and the other hears two explosions, with a time interval of one second between them. If the velocity of sound in metal is $3630 \,m \,s ^{-1},$ the distance between the two boys is

Waves

Solution:

$t_{\text {air }}=\frac{d}{330 m / s } ; t_{\text {metal }}=\frac{d}{3630 m / s }$
$t_{\text {air }}-t_{\text {metal }}=1 ; $
$\Rightarrow d\left[\frac{1}{330}-\frac{1}{3630}\right]=1$
$\Rightarrow d=\frac{3630 \times 330}{(3630-330)}$
$=\frac{3630 \times 330}{3300} \simeq 363 \,m$