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Question

A 40 cm long wire having a mass 3.2 gm and area of cross section 1 mm2 is stretched between two supports 40.05 cm apart. In its fundamental mode, it vibrates with a frequency 100064 Hz. Find the Young's modulus of the wire.

A
1×109 N-m2
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B
2×109 N-m2
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C
3×109 N-m2
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D
4×109 N-m2
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Solution

The correct option is A 1×109 N-m2
As linear mass density is
μ=3.2 gm40.05 cm=3.2×10340.05×102324000 kg/m
For fundamental mode, l=λ2
λ=2l


We use, fundamental frequency
n0=Vλ=12lTμ
100064=12×40.05×102  T324000
[100064×2×40.05×102]2324000=T
(100064)2×4×1600×104×324000T
T108 N

From Hooke's law, Young's modulus can be calculated as
Y=T/AΔL/L=10×0.48×106×0.05×102=109 N/m2

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