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A material has Poisson’s ratio as 0.5. If a uniform rod of it is subjected by a longitudinal strain of 2 x 10^-3, then what would be the percent increase in volume?

(a) 2

(b) 0

(c) 9

(d) 10

The question was asked in an interview for job.

The above asked question is from Strength of Materials in division Strength of Materials and Construction Materials of Farm Machinery

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The correct answer is (b) 0

For explanation I would say: \(\frac{∆l}{l}=2*10^{-3} \)

σ = 0.5

σ = \(\frac{[(\frac{∆r}{r})]}{[(\frac{∆l}{l})]} \)

\(\frac{∆r}{r}\) = -0.5*10^-3*2=1*10^-3

Volume of rod = πr^2l

Fractional increase in volume = \(\frac{∆v}{v}=\frac{[∆(πr^2 l)]}{[πr^2 l]} = \frac{∆l}{l}+\frac{2∆r}{r} \)

= 2*10^-3+2(-1*10^-3)

= 0

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