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A wire of length $0.4 \mathrm{~m}$ stretched at both ends vibrates 250 times per second. If the length of the wire is increased by $0.1 \mathrm{~m}$ and the stretching force is reduced to $1 / 4^{\text {th }}$ of its original value then the new frequency is
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The correct answer is:
$100 \mathrm{~Hz}$
We have
$\mathrm{f}=\frac{1}{2 \mathrm{~L}} \sqrt{\frac{\mathrm{T}}{\mu}}$
Divide equation (ii) by (i), we get
$$
\Rightarrow \frac{250}{\mathrm{f}^{\prime}}=\frac{5}{4} \times 2 \Rightarrow \mathrm{f}^{\prime}=100 \mathrm{~Hz}
$$
$\mathrm{f}=\frac{1}{2 \mathrm{~L}} \sqrt{\frac{\mathrm{T}}{\mu}}$
Divide equation (ii) by (i), we get
$$
\Rightarrow \frac{250}{\mathrm{f}^{\prime}}=\frac{5}{4} \times 2 \Rightarrow \mathrm{f}^{\prime}=100 \mathrm{~Hz}
$$
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