Question

A steel rod of length $1\,{\text{m}}$ and radius $10\,{\text{mm}}$ is stretched by a force $100\,{\text{kN}}$ along its length. The stress produced in the rod is then (${Y_{steel}} = 2 \times {10^{11}}\,{\text{N}} \cdot {{\text{m}}^{ - 2}}$)
A.$3.18 \times {10^6}\,{\text{N}} \cdot {{\text{m}}^{ - 2}}$
B.$3.18 \times {10^7}\,{\text{N}} \cdot {{\text{m}}^{ - 2}}$
C.$3.18 \times {10^8}\,{\text{N}} \cdot {{\text{m}}^{ - 2}}$
D.$3.18 \times {10^9}\,{\text{N}} \cdot {{\text{m}}^{ - 2}}$

Answer 1

Use the formula for the longitudinal stress acting on the wire or steel rod. This formula gives the relation between the force acting on the rod or wire and the cross-sectional area of the wire or rod. Determine the cross-sectional area of the rod and substitute it in the equation for longitudinal stress.

Formulae used:
The formula for longitudinal stress in a wire or rod is given by
${\text{Longitudinal stress}} = \dfrac{F}{A}$ …… (1)
Here, $F$ is the normal force acting on the wire or rod and $A$ is the cross-sectional area of the wire or rod.
The area of the circle is given by
$A = \pi {R^2}$ …… (2)
Here, $A$ is the area of the circle and $R$ is the radius of the circle.

Complete step by step answer:
We have given that the length of the steel rod is $1\,{\text{m}}$ and its radius is $10\,{\text{mm}}$.
$L = 1\,{\text{m}}$
$R = 10\,{\text{mm}}$
The force applied on the steel rod is $100\,{\text{kN}}$.
$F = 100\,{\text{kN}}$
Convert the unit of the radius of the steel rod in the SI system of units.
$R = \left( {10\,{\text{mm}}} \right)\left( {\dfrac{{{{10}^{ - 3}}\,{\text{m}}}}{{1\,{\text{mm}}}}} \right)$
$\Rightarrow R = 0.01\,{\text{m}}$
Hence, the radius of the steel rod is $0.01\,{\text{m}}$.
Convert the unit of the force applied on the steel rod in the SI system of units.
$F = \left( {100\,{\text{kN}}} \right)\left( {\dfrac{{{{10}^3}}}{{1\,{\text{k}}}}} \right)$
$\Rightarrow F = {10^5}\,{\text{N}}$
Hence, the force applied on the steel rod is ${10^5}\,{\text{N}}$.
We can calculate the cross-sectional area of the steel rod using equation (2).
Substitute $3.14$ for $\pi$ and $0.01\,{\text{m}}$ for $R$ in equation (2).
$A = \left( {3.14} \right){\left( {0.01\,{\text{m}}} \right)^2}$
$\Rightarrow A = 3.14 \times {10^{ - 4}}\,{{\text{m}}^2}$
Hence, the cross-sectional area of the steel rod is $3.14 \times {10^{ - 4}}\,{{\text{m}}^2}$.
Since the steel rod is stretched, the length of the steel rod increases. Hence, the stress acting on the steel rod is longitudinal stress.The longitudinal stress on the steel rod can be determined using equation (1).
Substitute ${10^5}\,{\text{N}}$ for $F$ and $3.14 \times {10^{ - 4}}\,{{\text{m}}^2}$ for $A$ in equation (1).
${\text{Longitudinal stress}} = \dfrac{{{{10}^5}\,{\text{N}}}}{{3.14 \times {{10}^{ - 4}}\,{{\text{m}}^2}}}$
$\therefore{\text{Longitudinal stress}} = 3.18 \times {10^8}\,{\text{N}} \cdot {{\text{m}}^{ - 2}}$

Therefore, the stress produced in the rod is $3.18 \times {10^8}\,{\text{N}} \cdot {{\text{m}}^{ - 2}}$.Hence, the correct option is C.

Note: The units of the radius of the steel rod and the force acting on the steel rod are not in the SI system of units. Hence, the students should not forget to convert the units of the radius of the steel rod and the force acting on it in the SI system of units. Otherwise, the final answer will not be correct.