A material has Poisson ratio $$0.5$$. If a rod of material has a longitudinal strain $$2$$ × $$10-3$$, the percentage change in volume is :

Question

A material has Poisson ratio $$0.5$$. If a rod of material has a longitudinal strain $$2$$ × $$10-3$$, the percentage change in volume is :

Infinity Learn · Accepted Answer

Here Poisson's ratio $$=$$ ∆rr∆llIn this case ∆ll $$=$$ $$2$$×$$10-3$$ and σ $$=$$ $$0.5$$    ∆rr $$=$$ σ∆ll           $$=$$ $$2$$×$$10-3$$×$$5.0$$ $$=$$ $$10-3Let$$ V be the initial volume and (V$$ $$+$$ ∆$$V) be the final volume, then          V $$=$$ π$$r2l$$          V $$+$$ ∆V $$=$$ π(r-$$∆$$r)2(l+$$∆$$l)          ∆V $$=$$ π$$r2$$∆$$l-2$$$$π$$r l ∆$$r(Subtracting$$ and $$simplifying)$$∆VV $$=$$ ∆$$ll-2$$∆rr         $$=$$ $$2$$ ×$$10-3-2$$×$$10-3$$ $$=$$ $$0Correct$$ choice is (4).

A material has Poisson ratio 0.5. If a rod of material has a longitudinal strain $2 \times 10^{- 3}$ , the percentage change in volume is :

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A material has Poisson ratio 0.5. If a rod of material has a longitudinal strain 2 × 10-3, the percentage change in volume is :

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A material has Poisson ratio 0.5. If a rod of material has a longitudinal strain $2 \times 10^{- 3}$ , the percentage change in volume is :