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The total length of a sonometer wire between fixed ends is 110 cm. Two bridges are placed to divide the length of wire in ratio 6 : 3 : 2. The tension in the wire is 400 N and the mass per unit length is 0.01 kg/m. What is the minimum common frequency with which three parts can vibrate?

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500 Hz

1100 Hz

166 Hz

1000 Hz

answer is B.

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Detailed Solution

$n = \frac{1}{2 L} \sqrt{\frac{T}{μ}} = \frac{1}{2 \times L} \sqrt{\frac{400}{0.01}} = \frac{10}{L}$

$n_{1} = α_{1} (\frac{10}{0.06}) = α_{1} \frac{1000}{6}$

$α_{1} = 6 \Rightarrow n_{1} = 1000 Hz$

$n_{2} = α_{2} (\frac{10}{0.03}) = α_{2} \frac{1000}{3}$

$α_{2} = 3 \Rightarrow n_{2} = 1000 Hz$

$n_{3} = α_{3} (\frac{10}{0.02}) = α_{3} \frac{1000}{2}$

$α_{3} = 2 \Rightarrow n_{3} = 1000 Hz$

Minimum common frequency is 1000 Hz.

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