AP EAMCET · PHYSICS · Mechanical Properties of Solids
When a wire of length \(10 \mathrm{~m}\) is subjected to a force of \(100 \mathrm{~N}\) along its length, the lateral strain produced is \(0.01 \times 10^{-3} \mathrm{~m}\). The Poisson's ratio was found to be 0.4 . If the area of cross-section of wire is \(0.025 \mathrm{~m}^2\), its Young's modulus is
- A \(1.6 \times 10^8 \mathrm{~N} / \mathrm{m}^2\)
- B \(2.5 \times 10^{10} \mathrm{~N} / \mathrm{m}^2\)
- C \(1.25 \times 10^{11} \mathrm{~N} / \mathrm{m}^2\)
- D \(16 \times 10^9 \mathrm{~N} / \mathrm{m}^2\)
Answer & Solution
Correct Answer
(A) \(1.6 \times 10^8 \mathrm{~N} / \mathrm{m}^2\)
Step-by-step Solution
Detailed explanation
Poisson's ratio \(=\frac{\text { lateral strain }}{\text { longitudinal strain }}\) \(\begin{aligned} \text { ie, } & 0.4 & =\frac{0.01 \times 10^{-3}}{\Delta L / L} \\ \text { or } & \frac{L}{\Delta L} & =\frac{0.4}{0.01 \times 10^{-3}}=4 \times 10^4\end{aligned}\) Young's…
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