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

Step 1: Calculate the Force
Force is given by:
\[F = mg\]
Here, $m = 4.0 \text{ kg}, g = 9.81 \text{ m/s}^2$:
\[F = 4.0 \times 9.81 = 39.24 \text{ N}\]
{Step 2: Calculate the Cross-Sectional Area}
Radius of the wire, $r = 2.0 \text{ mm} = 2.0 \times 10^{-3} \text{ m}$:
\[A = \pi r^2 = \pi (2.0 \times 10^{-3})^2 = 1.256 \times 10^{-5} \text{ m}^2\]
{Step 3: Calculate the Tensile Stress}
Tensile stress is given by:
\[\sigma = \frac{F}{A} = \frac{39.24}{1.256 \times 10^{-5}} \approx 62.4 \text{ MPa}\]