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  • Figure shows a graph of the extension of a wire of length <img alt="\mathrm

Figure shows a graph of the extension \mathrm{(\Delta l)} of a wire of length \mathrm{1 \mathrm{~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 \mathrm{10^{-6} \mathrm{~m}^2}, the Young's modulus of the material of the wire is

                     

 

Option: 1

\mathrm{2 \times 10^{11} \mathrm{~N} / \mathrm{m}^2}


Option: 2

\mathrm{2 \times 10^{-11} \mathrm{~N} / \mathrm{m}^2}


Option: 3

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


Option: 4

\mathrm{3 \times 10^{-12} \mathrm{~N} / \mathrm{m}^2}


Answers (1)

best_answer

Stress \mathrm{=\frac{\text { force }}{\text { area }}=\frac{W}{A} \cdot Strain =\frac{\Delta l}{l}.} Young's modulus is given by

\mathrm{Y=\frac{\text { stress }}{\text { strain }}=\frac{W / A}{\Delta l / l}=\frac{W}{\Delta l} \times \frac{l}{A}}                    (i)

\mathrm{\text { Now, slope of graph is } \frac{\Delta l}{W}=\frac{4 \times 10^{-4}}{80} \mathrm{~m} / \mathrm{N} \text {. Using }}

\mathrm{\text { this in (i), we get (given } l=1 \mathrm{~m} \text { and } A=10^{-6} \mathrm{~m}^2 \text { ) }}

\mathrm{Y=\frac{80}{4 \times 10^{-4}} \times \frac{1}{10^{-6}}=2 \times 10^{11} \mathrm{~N} / \mathrm{m}^2}

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avinash.dongre

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