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A closed pipe of length $300 \mathrm{~cm}$ contains some sand. A speaker is connected at one of its ends. The frequency of the speaker at which the sand will arrange itself in 20 equidistant piles is close to (velocity of sound is $300 \mathrm{~m} / \mathrm{s}$ )
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$1 \mathrm{kHz}$
$20 \frac{\lambda}{2}=300$
$\lambda=30 \mathrm{~cm}=0.3 \mathrm{~cm}$
$\mathrm{v}=v \lambda$
$\mathrm{v}=\frac{300}{0.3}=1000 \mathrm{~Hz}$
$\mathrm{v}=1 \mathrm{kHz}$
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