Question

If the given graph shows the load (W) attached to and the elongation ($\Delta l$) produced in a wire of length 1 meter and cross-sectional area 1 mm$^2$, then the Young's modulus of the material of the wire is

• A. \(20 \times 10^{10}\, \text{N/m}^2\)
• B. \(2 \times 10^{10}\, \text{N/m}^2\)
• C. \(10 \times 10^{10}\, \text{N/m}^2\)
• D. \(4 \times 10^{10}\, \text{N/m}^2\)

Hint:

Use the slope of the load vs elongation graph along with wire dimensions to calculate Young's modulus.

Answer 1

The Correct Option is B

Young's modulus \(Y\) is defined as: \[ Y = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{\Delta l / l} = \frac{F l}{A \Delta l} \] From the graph: When \(\Delta l = 5\, \text{mm} = 5 \times 10^{-3}\, \text{m}\), load \(W = 100\, \text{N}\)
Length \(l = 1\, m\), cross-sectional area \(A = 1\, \text{mm}^2 = 1 \times 10^{-6}\, \text{m}^2\)
Calculate: \[ Y = \frac{100 \times 1}{1 \times 10^{-6} \times 5 \times 10^{-3}} = \frac{100}{5 \times 10^{-9}} = 2 \times 10^{10}\, \text{N/m}^2 \]