Question

Figure represents the extension (\(\Delta l\)) of a wire of length \(l\), suspended from the ceiling of the room at one end and with a load \(W\) connected to the other end. If the cross-sectional area of the wire is \(10^{-5}\,\text{m}^2\) then the Young’s modulus of the wire is _____ N/m\(^2\).

• A. \(1.0\times10^{11}\)
• B. \(2.0\times10^{10}\)
• C. \(1.0\times10^{10}\)
• D. \(2.0\times10^{11}\)

Answer 1

The Correct Option is A

Concept: Young’s modulus is given by \[ Y=\frac{\text{stress}}{\text{strain}} \] \[ Y=\frac{F/A}{\Delta l/l} \] \[ Y=\frac{Fl}{A\Delta l} \] Step 1: {Extract values from graph.} From the graph: \[ W=60\,\text{N} \] \[ \Delta l = 6\times10^{-4}\,\text{m} \] Step 2: {Substitute in formula.} Given \[ A=10^{-5}\,\text{m}^2 \] \[ Y=\frac{60\times l}{10^{-5}\times6\times10^{-4}} \] Simplifying gives \[ Y=1.0\times10^{11}\,\text{N/m}^2 \]