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Question

A material has Poisson's ratio 0.50. If a uniform rod of it suffers a longitudinal strain of 2×103, then the percentage change in volume is

A
0.6
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B
0.4
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C
0.2
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D
zero
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Solution

The correct option is C zero
Given: dLL=2×103
Poisson's ratio, σ=drrdLL
where, r and L are the radius and length of the rod respectively
0.5=drr2×103

drr=103

Volume of the rod, V=πr2L
Differentiating, we get, dV=π(r2dL+2Lrdr)

dVV×100=π(r2dL+2rLdr)πr2L×100=(dLL+2drr)×100

dVV×100=[2×103+2(103)]×100=0

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