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A thin wire of length of $99 \mathrm{~cm}$ is fixed at both ends as shown in the figure. The wire is kept under a tension and is divided into three segments of lengths $l_1, l_2$ and $l_3$ as shown in figure. When the wire is made to vibrate, the segments vibrate respectively with their fundamental frequencies in the ratio $1: 2: 3$. Then, the lengths $l_1, l_2$ and $l_3$ of the segments respectively are (in cm)
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Verified Answer
The correct answer is:
$54,27,18$
Ratio of fundamental frequency
$$
\left(n_1: n_2: n_3\right)=1: 2: 3
$$
$$
\begin{aligned}
l_1: l_2: l_3 & =\frac{1}{n_1}: \frac{1}{n_2}: \frac{1}{n_3} \\
& =\frac{1}{1}: \frac{1}{2}: \frac{1}{3}=6: 3: 2 \\
l_1 & =\frac{6 \times 99}{11}=54 \\
l_2 & =\frac{3 \times 99}{11}=27 \\
l_3 & =\frac{2 \times 99}{11}=18
\end{aligned}
$$
$$
\left(n_1: n_2: n_3\right)=1: 2: 3
$$
$$
\begin{aligned}
l_1: l_2: l_3 & =\frac{1}{n_1}: \frac{1}{n_2}: \frac{1}{n_3} \\
& =\frac{1}{1}: \frac{1}{2}: \frac{1}{3}=6: 3: 2 \\
l_1 & =\frac{6 \times 99}{11}=54 \\
l_2 & =\frac{3 \times 99}{11}=27 \\
l_3 & =\frac{2 \times 99}{11}=18
\end{aligned}
$$
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