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Physics Question on Hooke's Law

The adjacent graph shows the extension (Δl)(\Delta l) of a wire of length 1 m suspended from the top of a roof at one end with a load W connected to the other end. If the cross-sectional area of the wire is 10−6m2,10^{-6} m^2 , calculate the Young's modulus of the material of the wire.

A

2×1011 Nm−22 \times 10^{11} \, Nm^{-2}

B

2×10−11 Nm−22 \times 10^{-11} \, Nm^{-2}

C

3×10−12 Nm−23 \times 10^{-12} \, Nm^{-2}

D

2×10−13 Nm−22 \times 10^{-13} \, Nm^{-2}

Answer

2×1011 Nm−22 \times 10^{11} \, Nm^{-2}

Explanation

Solution

From the graph l=10−4m,F=20 Nl = 10^{-4}m ,F = 20 \, N \hspace35mm A = 10^{-6} m^2 , L = 1 \, m \therefore \hspace30mm Y = \frac{FL}{Al} \hspace40mm = \frac{20 \times 1}{10^{-6} \times 10^{-4}} \hspace40mm = 20 \times 10^{10} = 2 \times 10^{11} \, Nm^{-2}