A tuning fork resonates with a sonometer wire of length $1 \mathrm{~m}$ stretched with a tension of $6 \mathrm{~N}$. When the tension in the wire is changed to $54 \mathrm{~N}$, the same tuning fork produces $12$ beats per second with it. The frequency of the tuning fork is $\mathrm{Hz}$.
JEE MAIN 2024, Diffcult
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$\mathrm{f}=\frac{1}{2 L} \sqrt{\frac{T}{\mu}}$
$\mathrm{f}_1=\frac{1}{2} \sqrt{\frac{6}{\mu}} \mathrm{f}_2=\frac{1}{2} \sqrt{\frac{54}{\mu}}$
$\frac{\mathrm{f}_1}{\mathrm{f}_2}=\frac{1}{3} \mathrm{f}_2-\mathrm{f}_1=12$
$\mathrm{f}_1=6 \mathrm{HZ}$
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