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A sonometer wire $49 \mathrm{~cm}$ long is in unison with a tuning fork of frequency ' $n$ '. If the length of the wire is decreased by $1 \mathrm{~cm}$ and it is vibrated with the same tuning fork, 6 beats are heard per second. The value of ' $n$ ' is
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The correct answer is:
$288 Hz$
$\begin{array}{ll}
& \mathrm{n}_1 \mathrm{~L}_1=\mathrm{n}_2 \mathrm{~L}_2 \\
\therefore & \mathrm{n}_1 \times 49=\mathrm{n}_2 \times 48 \\
\therefore & \mathrm{n}_2=\frac{49}{48} \mathrm{n}_1
\end{array}$
Given that, beat frequency, $\left|\mathrm{n}_1-\mathrm{n}_2\right|=6$
$\begin{array}{ll}
\therefore & \left|\mathrm{n}_1-\frac{49}{48} \mathrm{n}_1\right|=6 \\
\therefore & \frac{\mathrm{n}_1}{48}=6 \\
\therefore & \mathrm{n}_1=288 \mathrm{~Hz}
\end{array}$
& \mathrm{n}_1 \mathrm{~L}_1=\mathrm{n}_2 \mathrm{~L}_2 \\
\therefore & \mathrm{n}_1 \times 49=\mathrm{n}_2 \times 48 \\
\therefore & \mathrm{n}_2=\frac{49}{48} \mathrm{n}_1
\end{array}$
Given that, beat frequency, $\left|\mathrm{n}_1-\mathrm{n}_2\right|=6$
$\begin{array}{ll}
\therefore & \left|\mathrm{n}_1-\frac{49}{48} \mathrm{n}_1\right|=6 \\
\therefore & \frac{\mathrm{n}_1}{48}=6 \\
\therefore & \mathrm{n}_1=288 \mathrm{~Hz}
\end{array}$
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