TS EAMCET · Physics · Mechanical Properties of Solids
A tension of \(22 \mathrm{~N}\) is applied to a copper wire of cross-sectional area \(0.02 \mathrm{~cm}^2\) Young's modulus of copper is \(1.1 \times 10^{11} \mathrm{~N} / \mathrm{m}^2\) and Poisson's ratio 0.32 . The decrease in cross-sectional area will be
- A \(1.28 \times 10^{-6} \mathrm{~cm}^2\)
- B \(1.6 \times 10^{-6} \mathrm{~cm}^2\)
- C \(2.56 \times 10^{-6} \mathrm{~cm}^2\)
- D \(0.64 \times 10^{-6} \mathrm{~cm}^2\)
Answer & Solution
Correct Answer
(D) \(0.64 \times 10^{-6} \mathrm{~cm}^2\)
Step-by-step Solution
Detailed explanation
Young's modulus of materials \(Y=\frac{F \times l}{A \Delta l}, \frac{\Delta l}{l}=\frac{F}{A Y}\) \(=\frac{22}{0.02 \times 10^{-4} \times 1.1 \times 10^{11}}=10^{-4}\) Poisson's ratio, \(\sigma=\frac{\frac{\Delta l}{l}}{\frac{\Delta r}{r}}\)…
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