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The graph shows the extension of a wire of length 1.0 m suspended from the top of a roof at one end and with a load W connected to the other end. If the cross-sectional area of the wire is $10^{- 6} m^{2}$ , calculate the Young’s modulus of the material of the wire

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$2 \times 10^{11} N / m^{2}$

$2 \times 10^{10} N / m^{2}$

$3 \times 10^{12} N / m^{2}$

$2 \times 10^{13} N / m^{2}$

answer is A.

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Detailed Solution

Y=(Fl/Ae)

$Y = \frac{(\frac{80}{10^{- 6}})}{4 \times 10^{- 4}} = 2 \times 10^{11}$

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