A pipe’s lower end is immersed in water such that the length of air column from the top open end has a certain length $25\,\, cm$. The speed of sound in air is $350 \,\,m/s$. The air column is found to resonate with a tuning fork of frequency $1750 \,\,Hz$. By what minimum distance should the pipe be raised in order to make the air column resonate again with the same tuning fork ... $cm$ ?
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Given: speed of sound in air, $v=350 \mathrm{m} / \mathrm{s}$
For one sided closed pipe$:$
Frequency, $\nu_{n}=\frac{(2 n-1) v}{4 L} \quad$ where $n=1,2,3 \ldots \ldots$
When length of the pipe is $25 \mathrm{cm}$ OR $0.25 \mathrm{m}, \quad 1750=\frac{(2 n-1) 350}{4 \times 0.25}$
$\Longrightarrow n=3$
Now let the length of the pipe after raising be $L m$
For minimum increase in length of air column, $n=3+1=4$
$\Longrightarrow 1750=\frac{(2 \times 4-1) 350}{4 \times L}$ gives $L=0.35 \mathrm{m}$
Thus the distance by which the pipe is raised $=0.35-0.25=0.1 m=10 c m$
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