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

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$\frac{9}{4}$

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(D) The frequency $n$ of a stretched string is given by the formula $n = \frac{1}{2l} \sqrt{\frac{T}{m}}$,where $l$ is the length,$T$ is the tension,and $m$ is the mass per unit length.From this,we have the relation $n \propto \frac{\sqrt{T}}{l}$.Let the initial frequency be $n_1 = n$ and the final frequency be $n_2 = 2n$.Let the initial length be $l_1 = l$ and the final length be $l_2 = \frac{3}{4}l$.Using the proportionality,we get $\frac{n_1}{n_2} = \frac{l_2}{l_1} \sqrt{\frac{T_1}{T_2}}$.Substituting the values: $\frac{n}{2n} = \frac{\frac{3}{4}l}{l} \sqrt{\frac{T_1}{T_2}}$.$\frac{1}{2} = \frac{3}{4} \sqrt{\frac{T_1}{T_2}}$.Squaring both sides: $\frac{1}{4} = \frac{9}{16} \frac{T_1}{T_2}$.$\frac{T_2}{T_1} = \frac{9}{16} 	imes 4 = \frac{9}{4}$.Thus,the tension must be changed by a factor of $\frac{9}{4}$.

In order to double the frequency of the fundamental note emitted by a stretched string,the length is reduced to $\frac{3}{4}$ of the original length and the tension is changed. The factor by which the tension is to be changed is

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