A material has Poisson’s ratio 0.5 and Young’s modulus $2 \times 10^{11} Pa$ . If a uniform rod of circular cross section suffers a lateral strain of $4 \times 10^{- 6}$ , then percentage change in its density will be

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2 %

8 %

0 %

4 %

answer is C.

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Detailed Solution

$σ = - \frac{∆ r / r}{∆ l / l} = 0.5 \frac{∆ r}{r} = - σ \frac{∆ l}{l} = - 0.5 \times 4 \times 10^{- 6} V = π r^{2} l \frac{∆ V}{V} = 2 \frac{∆ r}{r} + \frac{∆ l}{l} = - 4 \times 10^{- 6} + 4 \times 10^{- 6} = 0$

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