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Figure shows a graph of the extension (A) of a wire of length 1 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 10-6m2 , the Young's modulus of the material of the wire is x×1011N/m2. The value of x is

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answer is 2.

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Detailed Solution

Stress = force area =WA: Strain =Δll. Young's modulus is given byY= stress strain =W/AΔl/l=WΔl×lA ……..(i)Now, slope of graph is ΔlW=4×10−480m/N. Using this in i), we get (given l=1m and A=10−6m2 ) Y=804×10−4×110−6=2×1011N/m2

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