The tension in a wire is decreased by $19 \%$. The percentage decrease in frequency will be ......... $\%$
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(b)
Velocity in a wave $=\sqrt{\frac{T}{\mu}}$
Fundamental frequency of waves $\frac{v}{2 l}$
$f=\sqrt{\frac{T}{\mu}} \times \frac{1}{2 l} \quad \dots (i)$
If $T$ decreases by $19 \%$ value of $T$ will be $T-0.19 T$
Putting this value in $(i)$
$f^{\prime}=\sqrt{\frac{T}{\mu}} \frac{(1-0.19)^{1 / 2}}{2 l}$
$f^{\prime}=f\left(1-\frac{1}{2} \times 0.19\right)$
[Binomial theorem]
$f^{\prime}=f-0.1 f$
Hence, the frequency decreases by $0.1 f$ are $10 \%$ of initial value.
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