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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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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