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A string in a musical instrument is $50 \mathrm{~cm}$ long and its fundamental frequency is $800 \mathrm{~Hz}$. Keeping the tension applied to the string same, the change in the length to produce sound note of fundamental frequency $1000 \mathrm{~Hz}$ will be.
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$40 \mathrm{~cm}$
For stretched string $\mathrm{v} \propto \frac{1}{l}$
$\therefore \frac{\mathrm{v}_1}{\mathrm{v}_2}=\frac{l_2}{l_1} \Rightarrow l_2=\frac{800}{1000} \times 50=40 \mathrm{~cm}$
$\therefore \frac{\mathrm{v}_1}{\mathrm{v}_2}=\frac{l_2}{l_1} \Rightarrow l_2=\frac{800}{1000} \times 50=40 \mathrm{~cm}$
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