A string of length $1\, m$ and mass $5\, g$ is fixed at both ends. The tension in the string is $8.0\, N$. The string is set into vibration using an external vibrator of frequency $100\, Hz$. The separation between successive nodes on the string is close to .... $cm$
JEE MAIN 2019, Medium
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$\mathrm{f}=\frac{\mathrm{n}}{2 \ell} \sqrt{\frac{\mathrm{T}}{\mathrm{m}}}$
On solving, $n=5$
$5$ loops are formed in $1 \mathrm{m}$
Separation between successive nodes $=\frac{1}{5} \mathrm{m}=20 \mathrm{cm}$
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