A person in front of a mountain is beating a drum at the rate of $40$ per minute and hears no distinct echo. If the person moves $90 \,m$ closer to the mountain, he has to beat the drum at $60$ per minute to not hear any distinct echo. The speed of sound is .............. $ms^{-1}$
KVPY 2018, Diffcult
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(c)
As drummer does not hear any echo this means time between two successive wavefronts is equal to time in which a wavefront reaches back to drummer.
Distance covered by sound $=2 x$
If $v=$ speed of sound, then
$\frac{2 x}{v}=$ time interval between two successive wavefronts.
So, we have
In case I, $\frac{2 x}{v}=\frac{60}{40} \quad \dots(i)$
In case II, $\frac{2(x-90)}{v}=\frac{60}{60} \quad \dots(ii)$
Substituting for $x$ from Eq. $(i)$ in Eq. $(ii)$, we get
$2 x-180=v$
$\Rightarrow \frac{3}{2}-180=v$
$\Rightarrow v=360 \,ms ^{-1}$
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