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  4. A stretched wire of length 110 cm is divided into three segments whose frequencies are in ratio 1:2:3 . Their length must be

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Q. A stretched wire of length $110\, cm$ is divided into three segments whose frequencies are in ratio $1:2:3 $ . Their length must be

Punjab PMETPunjab PMET 2007Electromagnetic Waves

Solution:

Given, $l_{1}+l_{2}+l_{3}=110\, cm$
and $n_{1} l_{1}=n_{2} l_{2}=n_{3} l_{3}$
$n_{1}: n_{2}: n_{3}:: 1: 2: 3$
$\because \frac{n_{1}}{n_{2}}=\frac{1}{2}=\frac{l_{2}}{l_{1}}$
$\Rightarrow l_{2}=\frac{l}{2}$
and $\frac{n_{1}}{n_{3}}=\frac{1}{3}=\frac{l_{3}}{l_{1}}$
$\Rightarrow l_{3}=\frac{l}{3}$
$\therefore l_{1}+\frac{l}{2}+\frac{l_{1}}{3}=110$
So, $l_{1}=60\, cm, l_{2}=30\, cm , l_{3}=20\, cm$