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A tension of 20N applied to a copper wire of cross sectional area 0.01cm². Young's modulus of copper is 1.1×10¹¹ N/m² and poisson's ratio is 0.32.The decrease in cross sectional area of the wire is

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a

1.16 × 10^–6cm²

b

1.16 × 10^–5cm²

c

1.16 × 10^–4cm²

d

1.16 × 10^–3cm²

answer is A.

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Detailed Solution

The relationship between stress and strain contains factor 2 if Poisson ratio (v) is specified.

$\frac{ΔA}{A} = \frac{2}{Y} \frac{F}{A} v$

$∆ A = \frac{2 F}{Y} v = 2 \frac{20}{1.1 \times 10^{11} N / m^{2}} (0.32) = 1.16 \times 10^{- 6} {cm}^{2}$

Hence the correct answer is $1.16 \times 10^{- 6} {cm}^{2}$

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A tension of 20N applied to a copper wire of cross sectional area 0.01cm2. Young's modulus of copper is 1.1×1011 N/m2 and poisson's ratio is 0.32.The decrease in cross sectional area of the wire is