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# The length of sonometer wire AB is $100cm$, where should the two bridges be placed from A to divide the wire in 3 segments whose fundamental frequencies are in the ratio of $1:2:6$A. $30cm,\,90cm$B. $60cm,\,90cm$C. $40cm,\,80cm$D. $20cm,\,30cm$

Last updated date: 22nd Jul 2024
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Hint: Sonometer is basically used to study the relationship between frequency, tension, and linear mass density and length of a stretched string. A Sonometer is a device based on the principle of Resonance. It is used to verify the laws of vibration of stretched string and also to determine the frequency of a tuning fork.

Given,
The length of sonometer wire, $AB = 100cm$
Ratio of segments $= 1:2:6$
Let ${L_1},{L_2}\,and\,{L_3}$be the lengths of three segments.
Then
${L_1} + {L_2} + {L_3} = 100$ ……………...(1)
Also the laws of vibrations of stretched strings
${f_1}{L_1} = {f_2}{L_2} = {f_3}{L_3}$
Given
$\Rightarrow {f_1}:{f_2}:{f_3} = 1:2:6$
Therefore
$= {L_1}:2{L_2}:6{L_3}$
So
${L_2} = \dfrac{{{L_1}}}{2}$
${L_3} = \dfrac{{{L_1}}}{6}$
Put the value in equation (1) and we get
$\Rightarrow {L_1} + \dfrac{{{L_1}}}{2} + \dfrac{{{L_1}}}{6} = 100$
Simplify
$\Rightarrow \dfrac{{6{L_1} + 3{L_1} + {L_1}}}{6} = 100$
$\Rightarrow 10{L_1} = 100 \times 6$
$\Rightarrow {L_1} = \dfrac{{600}}{{10}}$
$\Rightarrow {L_1} = 60cm$
Now
$\Rightarrow {L_2} = \dfrac{{{L_1}}}{2}$
$\Rightarrow {L_2} = \dfrac{{60}}{2}$
$\Rightarrow {L_2} = 30cm$
Again
$\Rightarrow {L_3} = \dfrac{{{L_1}}}{6}$
$\Rightarrow {L_3} = \dfrac{{60}}{6}$
$\Rightarrow {L_3} = 10cm$
Now
The first bridge, ${L_1} = 60cm$
The second bridge, ${L_1} + {L_2}$
$= 60 + 30$
$= 90cm$
The second bridge is, ${L_1} + {L_2} = 90cm$
So the answer is (2) $60cm,\,90cm$.