A fork of frequency $256\, Hz$ resonates with a closed organ pipe of length $25.4\, cm$. If the length of pipe be increased by $2\, mm$, the number of beats/sec. will be
AIIMS 2014, Medium
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$n_{1}=256=\frac{v}{4 \ell_{1}}=\frac{v}{4 \times 25.4}$
$\therefore \mathrm{v}=256 \times 101.6 \mathrm{cm} / \mathrm{s}$
$\mathrm{n}_{2}=\frac{\mathrm{v}}{4 \ell_{2}}=\frac{256 \times 101.6}{4 \times 25.6}=254 \mathrm{Hz}$
No. of $beats/sec$ $=\mathrm{n}_{1}-\mathrm{n}_{2}=256-254=2$
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