Question

A load of \(4.0 \, \text{kg}\) is suspended from a ceiling through a steel wire of length \(20 \, \text{m}\) and radius \(2.0 \, \text{mm}\). The length of the wire increases by \(0.031 \, \text{mm}\) at equilibrium. The Young’s modulus of steel is: (Take \(g = 9.81 \, \text{m/s}^2\)):

• A. \(60 \times 10^9 \, \text{N/m}^2\)
• B. \(50 \times 10^9 \, \text{N/m}^2\)
• C. \(40 \times 10^9 \, \text{N/m}^2\)
• D. \(20 \times 10^9 \, \text{N/m}^2\)

Hint:

When calculating Young’s modulus, ensure stress and strain are computed accurately. Always use SI units for force, area, and length.

Answer 1

The Correct Option is C

First, stress is determined using \[\sigma = \frac{F}{A}\]. With $F = 4.0 \times 9.81 = 39.24 \text{ N}$ and area $A = \pi r^2$, where $r = 2.0 \times 10^{-3} \text{ m}$, the calculated area is \[A = \pi (2.0 \times 10^{-3})^2 = 1.256 \times 10^{-5} \text{ m}^2\]. This yields a stress of \[\sigma = \frac{39.24}{1.256 \times 10^{-5}} = 3.12 \times 10^{6} \text{ N/m}^2\].
Next, strain is calculated using \[\epsilon = \frac{\Delta L}{L}\]. Given $L = 20 \text{ m}$ and $\Delta L = 0.031 \text{ mm} = 0.031 \times 10^{-3} \text{ m}$, the strain is \[\epsilon = \frac{0.031 \times 10^{-3}}{20} = 1.55 \times 10^{-6}\].
Finally, Young's Modulus is computed as \[Y = \frac{\sigma}{\epsilon} = \frac{3.12 \times 10^{6}}{1.55 \times 10^{-6}} \approx 40 \times 10^{9} \text{ N/m}^2\]