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(B) The total length of the wire is $L = 110 \ cm = 1.1 \ m$. The ratio of the lengths is $6:3:2$. Let the lengths be $l_1, l_2, l_3$. Sum of parts $=

Vedclass · Accepted Answer

$1000$

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(B) The total length of the wire is $L = 110 \ cm = 1.1 \ m$. The ratio of the lengths is $6:3:2$. Let the lengths be $l_1, l_2, l_3$. Sum of parts $= 6+3+2 = 11$. $l_1 = (6/11) 	imes 110 = 60 \ cm = 0.6 \ m$. $l_2 = (3/11) 	imes 110 = 30 \ cm = 0.3 \ m$. $l_3 = (2/11) 	imes 110 = 20 \ cm = 0.2 \ m$. The fundamental frequency of a wire is given by $f = \frac{1}{2l} \sqrt{\frac{T}{\mu}}$,where $T = 400 \ N$ and $\mu = 0.01 \ kg/m$. $\sqrt{\frac{T}{\mu}} = \sqrt{\frac{400}{0.01}} = \sqrt{40000} = 200 \ m/s$. $f_1 = \frac{200}{2 	imes 0.6} = \frac{100}{0.6} = \frac{1000}{6} \ Hz$. $f_2 = \frac{200}{2 	imes 0.3} = \frac{100}{0.3} = \frac{1000}{3} \ Hz$. $f_3 = \frac{200}{2 	imes 0.2} = \frac{100}{0.2} = 500 \ Hz = \frac{1000}{2} \ Hz$. The common frequency is the Least Common Multiple $(LCM)$ of the frequencies. $LCM(\frac{1000}{6}, \frac{1000}{3}, \frac{1000}{2}) = \frac{LCM(1000, 1000, 1000)}{GCD(6, 3, 2)} = \frac{1000}{1} = 1000 \ Hz$.

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