A 40 cm long wire having a mass 3.2 gm and area of cross section 1mm2 is stretched between two supports 40.05cm apart. In its fundamental mode, it vibrates with a frequency 100064Hz. 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 A1×109 N-m2 As linear mass density is μ=3.2 gm40.05 cm=3.2×10−340.05×10−2≈324000 kg/m
For fundamental mode, l=λ2 ⇒λ=2l
We use, fundamental frequency n0=Vλ=12l√Tμ ⇒100064=12×40.05×10−2
⎷T324000 ⇒[100064×2×40.05×10−2]2324000=T ⇒(100064)2×4×1600×10−4×324000≈T ⇒T≈108 N
From Hooke's law, Young's modulus can be calculated as Y=T/AΔL/L=10×0.48×10−6×0.05×10−2=109 N/m2