The total length of sonometer wire between fixed ends is 110 cm. The two bridges are placed to divide the length of wires in ratio 6 : 3 : 2. The tension in the wire is 400N 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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1100 Hz

1000 Hz

166 Hz

100 Hz

answer is B.

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Detailed Solution

Total length = 110 cm
lengths are in the ratio 6 : 3 : 2 i.e., 60 cm, 30 cm and 20 cm

$\begin{array}{l} v = \frac{1}{2 L} \sqrt{\frac{T}{μ}} = \frac{1}{2 L} v \\ f_{1} = \frac{n_{1} v}{2 l_{1}} = \frac{n_{1} v}{12 x} = \frac{500}{3} n_{1} (x = 10 cm) \\ f_{2} = \frac{n_{2} v}{2 I_{2}} = \frac{n_{2} v}{6 x} = \frac{1000}{3} n_{2} & f_{3} = \frac{n_{3} v}{2 I_{3}} = \frac{n_{3} v}{4 x} = 500 n_{3} \end{array}$
For all frequencies to be equal n₁ = 6, n₂ = 3 & n₃ = 2 hence, f = 1000 Hz