A resonance tube is old and has jagged end. It is still used in the laboratory to determine velocity of sound in air. A tuning fork of frequency $512\,Hz$ produces first resonance when the tube is filled with water to a mark $11\,cm$ below a reference mark, near the open end of the tube. The experiment is repeated with another fork of frequency $256\,Hz$ which produces first resonance when water reaches a mark $27\,cm$ below the reference mark. The velocity of sound in air, obtained in the experiment, is close to .... $ms^{-1}$
JEE MAIN 2019, Diffcult
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$\lambda_{1}=4(11+\mathrm{e}) =\frac{\mathrm{v}}{512}$
$\lambda_{2}=4(27+e)=\frac{V}{256}$
$\frac{11+e}{27+e}=\frac{1}{2}$
$\Rightarrow \quad 22+2 e=27+e$
$\Rightarrow \quad e=5$
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