Waves

JEE Main level, numerical answer, medium

135 of 142

Question 135 of 142

The speed of a transverse wave passing through a string of length 50cm50cm and mass 10g10g is 60ms−160m{s}^{-1}. The area of cross-section of the wire is 2.0mm22.0{mm}^{2} and its Young's modulus is 1.2×1011Nm−21.2\times {10}^{11}N{m}^{-2}. The extension of the wire over its natural length due to its tension will be x×10−5mx\times {10}^{-5}m. The value of xx is _____ .

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Answer: 15

The wave speed in string is given by vw=Tμ{v}_{w}=\sqrt{\dfrac{T}{\mu }}, where, TT is tension and μ=ml\mu =\dfrac{m}{l} is linear mass density.
Putting the values, we have
⇒60=T10×10−3×0.5\Rightarrow 60=\sqrt{\dfrac{T}{10\times {10}^{-3}}\times 0.5}
⇒T=(60)2×10−20.5=72N\Rightarrow T=\dfrac{{\left(60\right)}^{2}\times {10}^{-2}}{0.5}=72N
The Young's modulus is given by Y=FlA∆lY=\dfrac{Fl}{A∆l}, here, AA is area.
From above the extension of the wire is Δl=TlAY=72×0.52×10−6×1.2×1011\Delta l=\dfrac{Tl}{AY}=\dfrac{72\times 0.5}{2\times {10}^{-6}\times 1.2\times {10}^{11}}=72×524×10−5=15×10−5m=\dfrac{72\times 5}{24}\times {10}^{-5}=15\times {10}^{-5}m.
Hence, the value of x=15x=15.

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