Question

A load of \(4.0 \, \text{kg}\) is suspended from a ceiling fan through a steel wire of radius \(2.0 \, \text{mm}\). The tensile stress developed in the wire when equilibrium is achieved is: (Take \(g = 9.81 \, \text{m/s}^2\)):

• A. 31.2 MPa
• B. 62.4 MPa
• C. 93.6 MPa
• D. 124.8 MPa

Hint:

For tensile stress calculations, always convert all measurements (e.g., radius) into SI units. Double-check the area formula for circular cross-sections, A = πr^2 , to avoid errors.

Answer 1

The Correct Option is B

First, the force is determined using the formula `\[F = mg\]`. Given $m = 4.0 \text{ kg}$ and $g = 9.81 \text{ m/s}^2$, the force is calculated as `\[F = 4.0 \times 9.81 = 39.24 \text{ N}\]`.
Next, the cross-sectional area is calculated. With a wire radius $r = 2.0 \text{ mm} = 2.0 \times 10^{-3} \text{ m}$, the area is found using `\[A = \pi r^2 = \pi (2.0 \times 10^{-3})^2 = 1.256 \times 10^{-5} \text{ m}^2\]`.
Finally, the tensile stress is computed. Using the formula `\[\sigma = \frac{F}{A}\]`, the stress is `\[\sigma = \frac{39.24}{1.256 \times 10^{-5}} \approx 62.4 \text{ MPa}\]`.