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When a metal wire is stretched by a load, the fractional change in its volume ∆V/ V is proportional to(a) ∆l/l(b) (∆l/l)2(c) √(∆l/l)(d) none of these. |
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Answer» The correct answer is (a) ∆l/l Explanation: When a metal wire is stretched by a load the fractional change in the transverse length is proportional to the fractional change in the longitudinal length. i.e., Δd/d = σ*Δl/l, where σ is the Poisson's ratio. Let the original cross-sectional area =A and final area a. Change in volume ΔV = a*Δl and the original volume V =Al Hence the fractional change in the volume ΔV/V = a*Δl/Al = (a/A)*(Δl/l) =(d'²/d²)(Δl/l) ={(d-Δd)²/d²}(Δl/l) ={(d²-2d*Δd)/d²}(Δl/l) [taking Δd² negligible] So, ΔV/V = {1-2*Δd/d}(Δl/l) ={1-2* σ*(Δl/l)}*(Δl/l) =Δl/l - 2σΔl²/l² →ΔV/V = Δl/l {Taking Δl² negligible} Hence the option (a). |
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