A tuning fork is vibrating at $250\, {Hz}$. The length of the shortest closed organ pipe that will resonate with the tuning fork will be ..... ${cm}$
(Take speed of sound in air as $340\, {ms}^{-1}$ )
JEE MAIN 2021, Medium
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$\frac{\lambda}{4}=\ell \Rightarrow \lambda=4 \ell$
${f}=\frac{{V}}{\lambda}=\frac{{V}}{4 \ell}$
$\Rightarrow 250=\frac{340}{4 \ell}$
$\Rightarrow \ell=\frac{34}{4 \times 25}=0.34\, {m}$
$\ell=34\, {cm}$
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