Mechanical Properties of Solids

JEE Main level, numerical answer, medium

47 of 118

Question 47 of 118

In an experiment to determine the Young's modulus, steel wires of five different lengths (1,2,3,41,2,3,4 and 55) but of same cross-section (2mm2)\left(2{mm}^{2}\right) were taken and curves between extension and load were obtained. The slope (extension/load) of the curves were plotted with the wire length and the following graph is obtained. If the Young's modulus of given steel wires is x×1011Nm−2x\times {10}^{11}N{m}^{-2}, then the value of xx is _____ .

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Answer: 2

We know that, Young's modulus is Y=FLA∆LY=\dfrac{FL}{A∆L}, where, FF is force or load acting on wire, AA is cross-section area, LL is length and ∆L∆L is extension in length.
Now, from given graph, the slope of the line is =ΔLWL=ΔLLW=1YA=\dfrac{\dfrac{\Delta L}{W}}{L}=\dfrac{\dfrac{\Delta L}{L}}{W}=\dfrac{1}{YA}
⇒Y=1(slope)A\Rightarrow Y=\dfrac{1}{\left(slope\right)A}
Thus, Young's modulus of the steel wires is Y=12×10−6(0.25×10−5)Y=\dfrac{1}{2\times {10}^{-6}\left(0.25\times {10}^{-5}\right)}
⇒Y=2×1011Nm−2\Rightarrow Y=2\times {10}^{11}N{m}^{-2}.
Therefore, x=2x=2.

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