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In order to double the frequency of the fundamental note emitted by a stretched string, the length is reduced to $\frac{3}{4}$ th of the original length and the tension is changed. The factor by which the tension is to be changed is
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The correct answer is:
$\frac{9}{4}$
Frequency of stretched string.
$\mathrm{n}=\frac{1}{2 l} \frac{\sqrt{T}}{m}$
$\begin{aligned} & n \propto \frac{\sqrt{T}}{l} \\ & \frac{n_1}{n_2}=\frac{l_2}{I_1} \sqrt{\frac{T_1}{T_2}} \Rightarrow \frac{n}{2 n}=\frac{\frac{3}{4}}{l} \sqrt{\frac{T_1}{T_2}} \\ & \frac{1}{2}=\frac{3}{4} \sqrt{\frac{T_1}{T_2}} \Rightarrow \sqrt{\frac{T_1}{T_2}}=\frac{2}{3} \\ & \frac{T_1}{T_2}=\frac{4}{9} \Rightarrow T_2=\frac{9}{4} T_1\end{aligned}$
$\therefore$ Tension should be changed by the factor $\frac{9}{4}$.
$\mathrm{n}=\frac{1}{2 l} \frac{\sqrt{T}}{m}$
$\begin{aligned} & n \propto \frac{\sqrt{T}}{l} \\ & \frac{n_1}{n_2}=\frac{l_2}{I_1} \sqrt{\frac{T_1}{T_2}} \Rightarrow \frac{n}{2 n}=\frac{\frac{3}{4}}{l} \sqrt{\frac{T_1}{T_2}} \\ & \frac{1}{2}=\frac{3}{4} \sqrt{\frac{T_1}{T_2}} \Rightarrow \sqrt{\frac{T_1}{T_2}}=\frac{2}{3} \\ & \frac{T_1}{T_2}=\frac{4}{9} \Rightarrow T_2=\frac{9}{4} T_1\end{aligned}$
$\therefore$ Tension should be changed by the factor $\frac{9}{4}$.
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