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A wire of density 9×103 kg/m3 is stretched between two Clamps 1 m apart and is subjected to an extension of 4.9×10-4 m. The lowest frequency of transverse vibration in the wire is (Y=9×1010 N/m2

a
40 Hz
b
35 Hz
c
30 Hz
d
25 Hz

detailed solution

Correct option is B

For wire ifM= mass, ρ= density ,A= Area of cross section V= volume, l= length, Δl= change in length Then mass per unit length m=Ml=Alρl=Aρ And Young's modulus of elasticity Y=T/AΔl/l⇒ T=YΔlAlHence lowest frequency of vibrationn=12lTm=12lYΔllAAρ=12lYΔllρ⇒ n=12×19×1010×4.9×10−41×9×103=35Hz

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