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← JEE Main 2022 · 27 Jul · Shift 2
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JEE Main 2022 · 27 Jul · Shift 2, Physics question 51

In an experiment to determine the Young's modulus, steel wires of five different lengths (1, 2, 3, 4, and 5 m) but of same cross section (2 mm2^2) 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×1011x\times10^{11} Nm−2^{-2}, then the value of xx is ______.

Extension/Load (×10⁻⁵ N⁻¹ m, axis label printed as ×10⁻⁵ Nm⁻²) against wire length (m): a straight line through the origin with points (1, 0.25), (2, 0.5), (3, 0.75), (4, 1.0), (5, 1.25).
Answer and worked solution

Official answer: 2

Slope. extensionload=LAY\frac{\text{extension}}{\text{load}}=\frac L{AY}, so the graph's slope is 1AY=0.25×10−5\frac1{AY}=0.25\times10^{-5}.

Young's modulus. Y=12×10−6×0.25×10−5=2×1011Y=\frac1{2\times10^{-6}\times0.25\times10^{-5}}=2\times10^{11} N m−2^{-2}.

Answer: 2

Topics:Hooke's law and elastic moduliYoung's modulus by Searle's method
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