Q.

A material has Poisson’s ratio 0.5 and Young’s modulus  2×1011Pa . If a uniform rod of circular cross section suffers a lateral strain of 4×106, then percentage change in its density will be

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a

4 %

b

2 %

c

0 %

d

8 %

answer is C.

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Detailed Solution

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σ=-r/rl/l=0.5  rr=-σll=-0.5×4×10-6 V = πr2l VV=2rr+ll           = -4×10-6 +4×10-6 = 0 

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