A metal wire is stretched by a load. Then the fractional change in its volume ΔVV is proportional to
A
ΔLL
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B
(ΔL2)2
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C
√ΔLL
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D
(ΔLL)3
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Solution
The correct option is AΔLL From the data given in the question, a load F acting on the metallic wire is as shown in figure.
It is given that, the wire will stretch. ∴ The length of wire will increase
We know, Poisson's ratio, e=−ΔddΔLL−−(1)
Volume of wire, V=A×L=πd2L4 where d is the diameter of the wire. ⇒ΔV=πd24ΔL+πL4×2dΔd ⇒ΔVV=ΔLL+2Δdd ⇒ΔVV=(1−2e)ΔLL
[∴using (1)]
Therefore, ΔVV∝ΔLL