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

A) $20\ cm\ ;30\ cm\ ;60\ cm$
B) $60\ cm\ ;30\ cm\ ;20\ cm$
C) $60\ cm\ ;20\ cm\ ;30\ cm$
D) $30\ cm\ ;60\ cm\ ;20\ cm$
Explanation

(b) $l_{1} + l_{2} + l_{3} = 110\ cm$ and $n_{1}l_{1} = n_{2}l_{2} = n_{3}l_{3}$
$n_{1}:n_{2}\ :n_{3} :: 1\ :2\ :3$
$\because\frac{n_{1}}{n_{2}} = \frac{1}{2} = \frac{l_{2}}{l_{1}} \Rightarrow l_{2} = \frac{l_{1}}{2}$ and $\frac{n_{1}}{n_{3}} = \frac{1}{3} = \frac{l_{3}}{l_{1}} \Rightarrow l_{3} = \frac{l_{1}}{3}$
$\because l_{1} + \frac{l_{1}}{2} + \frac{l_{1}}{3} = 110\ so\ l_{1} = 60cm,\ l_{2} = 30cm,\ l_{3} = 20cm$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.

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