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Question

The graph shows the extension (Δl) 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 106 m2, Young's modulus of the material of the wire is


A
2×1011 N/m2
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B
2×1011 N/m2
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C
3×1012 N/m2
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D
2×1013 N/m2
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Solution

The correct option is A 2×1011 N/m2
We know that, Y=FAΔll
where Y = Young's modulus of elasticity
Y=F×lA×Δl=WΔl×lA

At the point where W=20 N, Δl=1×104 m [from the graph]
Y=20104×1106
Y=2×1011 N/m2

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