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

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$30 c m; 60 c m; 20 c m$

$60 c m; 20 c m; 30 c m$

$60 c m; 30 c m; 20 c m$

$20 c m; 30 c m; 60 c m$

answer is B.

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Detailed Solution

$\begin{array}{l} l_{1} + l_{2} + l_{3} = 110 c m a n d n_{1} l_{1} = n_{2} l_{2} = n_{3} l_{3} \\ n_{1} : n_{2} : n_{3} : 1 : 2 : 3 \\ \frac{n_{1}}{n_{2}} = \frac{1}{2} = \frac{l_{2}}{l_{1}} \Rightarrow l_{2} = \frac{l_{1}}{l_{2}} a n d \frac{n_{1}}{n_{3}} = \frac{1}{3} = \frac{l_{3}}{l_{1}} \Rightarrow l_{3} = \frac{l_{1}}{3} \\ l_{1} + \frac{l_{1}}{2} + \frac{l_{1}}{3} = 110 s o l_{1} = 60 c m, l_{2} = 30 c m, l_{3} = 20 c m . \end{array}$