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(B) The frequency $n$ of a stretched string is given by $n = \frac{1}{2l} \sqrt{\frac{T}{\mu}}$,where $T$ is the tension,$l$ is the length,and $\mu$ i

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$30$

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(B) The frequency $n$ of a stretched string is given by $n = \frac{1}{2l} \sqrt{\frac{T}{\mu}}$,where $T$ is the tension,$l$ is the length,and $\mu$ is the mass per unit length.Since $n$ and $\mu$ are constant,we have $n \propto \frac{\sqrt{T}}{l}$,which implies $l \propto \sqrt{T}$.Given that the tension is increased by $69\%$,the new tension $T_2 = T_1 + 0.69 T_1 = 1.69 T_1$.Using the proportionality $l_2 / l_1 = \sqrt{T_2 / T_1}$,we get $l_2 / l_1 = \sqrt{1.69 T_1 / T_1} = \sqrt{1.69} = 1.3$.Therefore,$l_2 = 1.3 l_1 = l_1 + 0.3 l_1$.This represents an increase of $30\%$ in length.

The tension of a stretched string is increased by $69\%$. In order to keep its frequency of vibration constant,its length must be increased by .... $\%$

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