Mechanical Properties of Solids

JEE Main level, numerical answer, medium

64 of 118

Question 64 of 118

A wire of density 9×10–3kgcm–39\times {10}^{–3}kg{cm}^{–3} is stretched between two clamps 1m1m apart. The resulting strain in the wire is 4.9×10–44.9\times {10}^{–4}. The lowest frequency of the transverse vibrations in the wire (Young's modulus of wire Y=9×1010Nm–2Y=9\times {10}^{10}{Nm}^{–2} ), (to the nearest integer),_______

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Answer: 35

Fundamental frequency in the stringf=12ℓTμ=12ℓTρA=12ℓYΔℓρℓf=\dfrac{1}{2\ell }\sqrt{\dfrac{T}{\mu }}=\dfrac{1}{2\ell }\sqrt{\dfrac{T}{\rho A}}=\dfrac{1}{2\ell }\dfrac{\sqrt{Y\Delta \ell }}{\rho \ell }
f=12ℓYΔℓρℓf=\dfrac{1}{2\ell }\sqrt{\dfrac{Y\Delta \ell }{\rho \ell }}
=129×1010×4.9×10−49×10−3×10−6=35=\dfrac{1}{2}\sqrt{\dfrac{9\times {10}^{10}\times 4.9\times {10}^{-4}}{9\times {10}^{-3}\times {10}^{-6}}}=35

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