Question

A steel wire having a radius of $2.0mm$ , carrying a load of $4kg$ , is hanging from a ceiling. Given that $g = \left( {3.1\pi } \right)m{s^{ - 2}}$ , what will be the tensile stress that would be developed in the wire?
A. $4.8 \times {10^6}N{m^{ - 2}}$
B. $5.2 \times {10^6}N{m^{ - 2}}$
C. $6.2 \times {10^6}N{m^{ - 2}}$
D. $3.1 \times {10^6}N{m^{ - 2}}$

Answer 1

A tensile stress is created due to tension in any string or wire. It is the internal restoring force (in case of tensile stress it is the tension force) acting per unit area of the deformed string/wire. It is given by ${\text{Stress}} = \dfrac{{{\text{Restoring force}}}}{{{\text{Area}}}}$ .

Complete step by step answer:
As given in the question that a load of mass $m = 4kg$ is hanging on a wire of radius $2mm$ .
Let us first understand the term stress. It is basically occurs when a force acts on a body and the body gets deformed due to this force. It is the internal restoring force (in case of tensile stress it is the tension force) acting per unit area of the deformed string/wire. It is given by ${\text{Stress}} = \dfrac{{{\text{Restoring force}}}}{{{\text{Area}}}}$ . A tensile stress is created due to tension in any string or wire.
Now, we know that the restoring force in this case is the tension in the wire. So,
Tension, $T = mg = 4 \times 3.1 \times \pi {\text{ N}}$ (as given that $g = 3.14\pi {\text{ m}}{{\text{s}}^{ - 2}}$ )
Now, we find the cross section area of the wire by using the formula $A = \pi {r^2}$ .
So, area $A = \pi \times {\left( {2 \times {{10}^{ - 3}}} \right)^2} = \pi \times 4 \times {10^{ - 6}}{\text{ }}{{\text{m}}^2}$
Now, we can calculate the stress
${\text{Stress}} = \dfrac{{4 \times 3.1 \times \pi }}{{\pi \times 4 \times {{10}^{ - 6}}}} = 3.1 \times {10^6}{\text{ N}}{{\text{m}}^{ - 2}}$
Hence, option D is correct.

Note: Apart from the tensile stress, there are several more types of stress such as bulk stress, compressive stress and tangential or shearing stress. These differ mainly with the type of restoring forces and the type of body on which the force is applied. The restoring force always acts in the direction opposite to the direction of deforming or external force having magnitude equal to the external force.