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Step-by-Step Solution
Step 1: List the Given Data
• Mass of the load, $m = 4\,\text{kg}$
• Radius of the wire, $R = 2.0\,\text{mm} = 2.0 \times 10^{-3}\,\text{m}$
• Acceleration due to gravity, $g = 3.1\,\text{m/s}^2$
Step 2: Write the Formula for Tensile Stress
Tensile stress ($\sigma$) is defined as the ratio of the tensile force ($F$) to the cross-sectional area ($A$) of the wire. Mathematically,
$\sigma = \dfrac{F}{A}$
Here, the force $F$ is given by
$F = mg$
and the cross-sectional area of the wire, assuming it is circular, is
$A = \pi R^2$
Step 3: Substitute the Known Values
• Compute $F = mg = 4 \times 3.1 = 12.4\,\text{N}$
• Compute $A = \pi R^2 = \pi \times (2.0 \times 10^{-3})^2 = \pi \times 4.0 \times 10^{-6}\,\text{m}^2$
So the stress becomes:
$\sigma = \dfrac{12.4}{\pi \times 4.0 \times 10^{-6}} \,\text{N/m}^2$
Step 4: Final Calculation
Now, simplifying,
$\sigma = \dfrac{12.4}{4.0 \,\pi \times 10^{-6}} = \dfrac{12.4}{4 \times 10^{-6}} \times \dfrac{1}{\pi} \times \pi$
Noting that $\pi$ cancels out, it reduces to:
$\sigma = \dfrac{12.4}{4 \times 10^{-6}} = 3.1 \times 10^6 \,\text{N/m}^2$
Step 5: State the Final Answer
Therefore, the tensile stress developed in the wire is
$\boxed{3.1 \times 10^6\,\text{N/m}^2}$
