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A string is divided into three segments, so that the segment possesses fundamental frequencies in the ratio $1: 2: 3$. Then, the length of the segments are in the ratio
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The correct answer is:
$6: 3: 2$
Given, ratio of fundamental frequencies be $f_1: f_2: f_3=1: 2: 3$
Let ratio of length be $l_1: l_2: l_3$.
Since, $\quad f=\frac{1}{2 l} \sqrt{\frac{T}{\mu}}$
where, $f$ is frequency,
$l$ is length of string,
$T$ is tension
and $\mu$ is mass per unit length.
$\therefore \quad l \propto \frac{1}{f}$
So, $l_1: l_2: l_3=\frac{1}{f_1}: \frac{1}{f_2}: \frac{1}{f_3}$
$=\frac{1}{1}: \frac{1}{2}: \frac{1}{3}$
$=\frac{6}{6}: \frac{3}{6}: \frac{2}{6}$
$=6: 3: 2$
Let ratio of length be $l_1: l_2: l_3$.
Since, $\quad f=\frac{1}{2 l} \sqrt{\frac{T}{\mu}}$
where, $f$ is frequency,
$l$ is length of string,
$T$ is tension
and $\mu$ is mass per unit length.
$\therefore \quad l \propto \frac{1}{f}$
So, $l_1: l_2: l_3=\frac{1}{f_1}: \frac{1}{f_2}: \frac{1}{f_3}$
$=\frac{1}{1}: \frac{1}{2}: \frac{1}{3}$
$=\frac{6}{6}: \frac{3}{6}: \frac{2}{6}$
$=6: 3: 2$
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