AP EAMCET · PHYSICS · Mechanical Properties of Solids
A wire of length 100 cm and area of cross-section \(2 \mathrm{~mm}^2\) is stretched by two forces of each 440 N applied at the ends of the wire in opposite directions along the length of the wire. If the elongation of the wire is 2 mm , the Young's modulus of the material of the wire is
- A \(4.4 \times 10^{11} \mathrm{Nm}^{-2}\)
- B \(1.1 \times 10^{11} \mathrm{Nm}^{-2}\)
- C \(2.2 \times 10^{11} \mathrm{Nm}^{-2}\)
- D \(3.3 \times 10^{11} \mathrm{Nm}^{-2}\)
Answer & Solution
Correct Answer
(B) \(1.1 \times 10^{11} \mathrm{Nm}^{-2}\)
Step-by-step Solution
Detailed explanation
\(1=100 \mathrm{~cm}=100 \times 10^{-2} \mathrm{~m}, \mathrm{~A}=2 \mathrm{~mm}^2=2 \times 10^{-6} \mathrm{~m}^2\) \(\mathrm{F}=440 \mathrm{~N}, \delta \mathrm{l}=2 \mathrm{~mm}=2 \times 10^{-3} \mathrm{~m}\) \(\therefore\) Young's modulus of the wire is…
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