JEE Mains · Physics · STD 11- 8. mechanical properties of solids
Figure represents the extension (\(\Delta l\)) of a wire of length \(1\) meter, suspended from the ceiling of the room at one end with a load \(W\) connected to the other end. If the cross-sectional area of the wire is \(10^{-5}\) m\(^2\) then the Young's modulus of the wire is __________ N/m\(^2\).
- A \(1.0 \times 10^{11}\)
- B \(2.0 \times 10^{10}\)
- C \(1.0 \times 10^{10}\)
- D \(2.0 \times 10^{11}\)
Answer & Solution
Correct Answer
(C) \(1.0 \times 10^{10}\)
Step-by-step Solution
Detailed explanation
From the given graph, we can observe that for a load \(W = 20\) N, the extension is \(\Delta l = 2 \times 10^{-4}\) m. The formula for Young's modulus \(Y\) is: \(Y = \dfrac{\text{Stress}}{\text{Strain}} = \dfrac{W / A}{\Delta l / L} = \dfrac{W L}{A \Delta l}\) Given: Length of…
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