Question

A thin wire of length of $99\, cm$ 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 $I_{1}, I_{2}$ 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 $\left(n_{1}: n_{2}: n_{3}\right)=1: 2: 3$

$l_{1}: 1_{2}: l_{3}=\frac{1}{n_{1}}: \frac{1}{n_{2}}: \frac{1}{n_{3}}$
$=\frac{1}{1}: \frac{1}{2}: \frac{1}{3}=6: 3: 2$
$l_{1}=\frac{6 \times 99}{11}=54$
$l_{2}=\frac{3 \times 99}{11}=27$
$l_{3}=\frac{2 \times 99}{11}=18$