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Model Paper 3 — Science STD 9 — Question

Gujarat BoardEnglish MediumSTD 9ScienceModel Paper 33 Marks
Question
A person standing benween two vertical cliffs and $640 m$ away from the nearest cliff shouted. He heard the first echo ather $4$ seconds and the second echo $3$ seconds later. Calculate
$I$. the velocity of sound in air, and
$II.$ the distance between the cliffs.
✓

Answer

$i$. Let $P$ be the person standing between the cliffs $A$ and $B$ .
Let $s _1$ be a distance of nearest cliff $A$ from $P$ and $s _2$ the distance of second cliff $B$ from $P$.
The first echo is heard when sound reaches the person after being reflected from cliff $A$
Image
Given, $s _1= AP =640 m$
Time interval of first echo, $t _1=4$ seconds
From relation, $2 s_1=v t _1$, we have
The speed of sound, $v=\frac{2 s_1}{t_1}$
$ \frac{2 \times 640}{4} $
Therefore, Speed of sound in air, $v =320 m / s$
$ii$. The second echo is heard when the sound reaches the person after being reflected from the cliff $B$ .
Time interval of second echo, $t _2=4+3=7$ seconds
Therefore, From relation, $2 s_2=v t _2$,
We have, $\frac{v t_2}{2}$
$ \frac{320 \times 7}{2}$
$=1120 m $
Therefore, Distance between cliffs $A$ and $B ,$
$ s=s_1+s_2$
$=640+1120=1760 m $

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