Question

A thin wire of $99\text{cm}$ is fixed at both ends as shown in 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. If the wires are made to vibrate, with their fundamental frequencies respectively in the ratio $1:2:3$, then the lengths $l_1,l_2,l_3$ of the segments respectively are (in $\text{cm}$):

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

Answer 1

The correct option is C $54,27,18$

We know from law of lengths, $n\propto \frac{1}{l}$
$\Rightarrow l_1:l_2:l_3=\frac{1}{n_1}:\frac{1}{n_2}:\frac{1}{n_3}$

From the data given in the question,
$l_1:l_2:l_3=1:\frac{1}{2}:\frac{1}{3}=6:3:2$
Let common factor be '$a$'
Then, $6a+3a+2a=99\Rightarrow a=9$
$l_1=6a=6\times 9=54\text{cm}$,
$l_2=3a=3\times 9=27\text{cm}$
and
$l_3=2a=2\times 9=18\text{cm}$

Thus, option (c) is the correct answer.