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A steel wire of length 1m, mass 0.1kg and uniform cross sectional area 10^-6 m² is rigidly fixed at both ends. The temperature of the wire is lowered by 20°C. If the transverse waves are set up plucking the string in the middle, calculate the frequency of the fundamental mode of vibration.
$(Y = 2 x 10^{11} N / m^{2}, a = 1.21 x 10^{- 5} /^{o} C$

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22 Hz

33 Hz

44 Hz

11 Hz

answer is D.

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Detailed Solution

$n = \frac{1}{2 l} \sqrt{\frac{yAα ∆ θ_{l}}{m}} n = \frac{1}{2 x 1} \sqrt{\frac{2 x 10^{11} x 10^{- 6} x 1.21 x 10^{- 5} x 20}{0.1}} = \frac{1}{2} \sqrt{40 x 12.1} = \frac{2 x 11}{2} = 11 Hz$

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