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If man were standing unsymmetrically between parallel cliffs, claps his hands and starts hearing a series of echoes at intervals of $1 \mathrm{~s}$. If speed of sound in air is $340 \mathrm{~m} \mathrm{~s}^{-1}$, the distance between two cliffs would be
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Verified Answer
The correct answer is:
$510 \mathrm{~m}$
Let $x$ be distance of person from one cliff and $y$ be distance of person from $2^{\text {nd }}$ cliff. Let
$$
\begin{aligned}
y>x & \\
\therefore \quad x & +x=v \times t_1=340 \times 1=340 \\
x & =170 \mathrm{~m} \\
y+y & =v \times t_2=340 \times 2=680 \\
y & =340 \mathrm{~m} .
\end{aligned}
$$
Distance between two cliffs
$$
=x+y=170+340=510 \mathrm{~m}
$$
$$
\begin{aligned}
y>x & \\
\therefore \quad x & +x=v \times t_1=340 \times 1=340 \\
x & =170 \mathrm{~m} \\
y+y & =v \times t_2=340 \times 2=680 \\
y & =340 \mathrm{~m} .
\end{aligned}
$$
Distance between two cliffs
$$
=x+y=170+340=510 \mathrm{~m}
$$
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