Question

A load of 4.0 kg is suspended from a ceiling through a steel wire of radius 2.0 mm. Find the tensile stress developed in the wire when equilibrium is achieved. Take $g=3.1\pi m{s}^{-2}.$

• A. $3.1\times {10}^{4}N/m^2$
• B. $3.2\times {10}^{5}N/m^2$
• C. $3.1\times {10}^{6}N/m^2$
• D. $3.1\times {10}^{7}N/m^2$

Answer 1

The correct option is C $3.1\times {10}^{6}N/m^2$

Tension in the wire is
F = 4.0 $\times 3.1\pi N.$
The area of cross section is
$A=\pi r^2=\pi \times (2.0\times {10}^{-3}m{)}^2$
$=4.0\pi \times {10}^{-6}{m}^2.$
Thus, the tensile stress developed
$=\frac{F}{A}=\frac{4.0\times 3.1\pi }{4.0\pi \times {10}^{-6}}N{m}^{-2}$
$=3.1\times {10}^{6}N{m}^{-2}.$