Question

A thin wire of length of $99cm$ 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 $I_1,I_2$ and $I_3$ 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 $I_1,I_2$ and $l_3$ of the segments respectively are (in $cm$ )

• A. 27,54,18
• B. 18,27,54
• C. 54,27,18
• D. 27,9,14

Answer 1

Answer: (C)

Ratio of fundamental frequency $\left(n_1:n_2:n_3\right)=1:2:3$

$l_1:1_2:l_3=\frac{1}{n_1}:\frac{1}{n_2}:\frac{1}{n_3}$
$=\frac{1}{1}:\frac{1}{2}:\frac{1}{3}=6:3:2$
$l_1=\frac{6\times 99}{11}=54$
$l_2=\frac{3\times 99}{11}=27$
$l_3=\frac{2\times 99}{11}=18$