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(C) The fundamental frequency of a stretched string is given by $n = \frac{1}{2l} \sqrt{\frac{T}{m}}$.From this,we can see that $n \propto \frac{\sqrt

Vedclass · Accepted Answer

$40$

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(C) The fundamental frequency of a stretched string is given by $n = \frac{1}{2l} \sqrt{\frac{T}{m}}$.From this,we can see that $n \propto \frac{\sqrt{T}}{l}$.Let the initial length be $l$ and initial tension be $T$. The final length is $l' = l - \frac{x}{100}l = l(1 - \frac{x}{100})$ and the final tension is $T' = T + 0.44T = 1.44T$.The ratio of the final frequency $n'$ to the initial frequency $n$ is given as $\frac{n'}{n} = \frac{2}{1}$.Using the proportionality relation: $\frac{n'}{n} = \sqrt{\frac{T'}{T}} 	imes \frac{l}{l'}$.Substituting the values: $\frac{2}{1} = \sqrt{\frac{1.44T}{T}} 	imes \frac{l}{l(1 - \frac{x}{100})}$.$2 = \sqrt{1.44} 	imes \frac{1}{1 - \frac{x}{100}}$.$2 = 1.2 	imes \frac{100}{100 - x}$.$2(100 - x) = 120$.$200 - 2x = 120$.$2x = 80$.$x = 40$.

If the length of a stretched string is shortened by $x \%$ and the tension is increased by $44 \%$,then the ratio of the final and initial fundamental frequencies is $2:1$. The value of $x$ is:

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