AP EAMCET · PHYSICS · Mechanical Properties of Solids
A tension of \(20 \mathrm{~N}\) is applied to a copper wire of cross sectional area \(0.01 \mathrm{~cm}^2\), Young's modulus of copper is \(1.1 \times 10^{11} \mathrm{~N} / \mathrm{m}^2\) and Poisson's ratio is 0.32 . The decrease in cross sectional area of the wire is
- A \(1.16 \times 10^{-6} \mathrm{~cm}^2\)
- B \(1.16 \times 10^{-5} \mathrm{~m}^2\)
- C \(1.16 \times 10^{-4} \mathrm{~m}^2\)
- D \(1.16 \times 10^{-3} \mathrm{~cm}^2\)
Answer & Solution
Correct Answer
(A) \(1.16 \times 10^{-6} \mathrm{~cm}^2\)
Step-by-step Solution
Detailed explanation
Given, \(\sigma=0.32, F=20 \mathrm{~N}\) \[ A=0.01 \mathrm{~cm}^2=0.01 \times 10^{-3} \mathrm{~m} \] and \(Y=1-1 \times 10^{11} \mathrm{~N} / \mathrm{m}^2\) We know that…
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