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(D) Let the initial frequency of each wire be $n$. The frequency of a stretched wire is given by $n = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$,which implies

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$300$

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(D) Let the initial frequency of each wire be $n$. The frequency of a stretched wire is given by $n = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$,which implies $n \propto \sqrt{T}$.Taking the derivative,we get $\frac{dn}{n} = \frac{1}{2} \frac{dT}{T}$.Given the percentage change in tension is $1\%$,so $\frac{dT}{T} = 0.01$.The change in frequency is $\Delta n = \frac{1}{2} 	imes n 	imes \frac{dT}{T} = \frac{1}{2} 	imes n 	imes 0.01 = 0.005n$.We are given that $3$ beats are heard in $2 \,s$,so the beat frequency is $\Delta n = \frac{3}{2} = 1.5 \,Hz$.Equating the two expressions for $\Delta n$: $0.005n = 1.5$.Solving for $n$: $n = \frac{1.5}{0.005} = 300 \,Hz$ or $300 \,s^{-1}$.

Two perfectly identical wires are in unison. When the tension in one wire is increased by $1\%$,then on sounding them together $3$ beats are heard in $2 \,s$. The initial frequency of each wire is .... $\text{s}^{-1}$.

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