AP EAMCET · PHYSICS · Mechanical Properties of Solids
A copper wire of cross-sectional area \(0.01 \mathrm{~cm}^2\) is under a tension of \(22 \mathrm{~N}\). The decrease in the cross-sectional area is
(Young modulus \(=1.1 \times 10^{11} \mathrm{Nm}^{-2}\), Poisson's ratio \(=0.32\) )
- A \(0.128 \times 10^{-6} \mathrm{~cm}^2\)
- B \(128 \times 10^{-6} \mathrm{~cm}^2\)
- C \(12.8 \times 10^{-6} \mathrm{~cm}^2\)
- D \(1.28 \times 10^{-6} \mathrm{~cm}^2\)
Answer & Solution
Correct Answer
(D) \(1.28 \times 10^{-6} \mathrm{~cm}^2\)
Step-by-step Solution
Detailed explanation
Young's modulus, \[ \begin{aligned} Y & =\frac{F / A}{\Delta l / l} \\ \frac{\Delta l}{l} & =\frac{F}{Y A} \end{aligned} \] where, \(\frac{\Delta l}{l}=\) longitudinal strain Given, \(F=22 \mathrm{~N}, Y=1.1 \times 10^{11} \mathrm{~N}-\mathrm{m}^2\),…
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