First slide

Elastic Modulie

Question

Increase in length of a wire on stretching is 0.025%. If its poisson's ratio is 0.4, percentage decrease in diameter is:

Easy

A

0.03%
B

0.02%
C

0.01%
D

0.05%

Solution

$\frac{∆ l}{l} = \frac{0.025}{100} = 2.5 \times 10^{- 4} (- \frac{∆ D}{D}) = σ \times (\frac{∆ l}{l}) = 0.4 \times 2.5 \times 10^{- 4} = 10^{- 4} (- \frac{∆ D}{D}) \times 100 = 0.01 %$

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