Question

In an experiment, brass and steel wires of length $1\;m$ each with areas of the cross-section $1m{m^2}$ are used. The wires are connected in series and one end of the combined wire is connected to a rigid support and the other end is subjected to elongation. The stress required to produce a net elongation of $0.2\;mm$ is,
[Given, Young’s modulus for steel and brass are, respectively $120 \times {10^9}N/{m^2}$ and $60 \times {10^9}N/{m^2}$ ]
(A) $1.8 \times {10^6}N/{m^2}$
(B) $0.2 \times {10^6}N/{m^2}$
(C) $4.0 \times {10^6}N/{m^2}$
(D) $1.2 \times {10^6}N/{m^2}$

Answer 1

Hint: - The relation between stress experienced by the elements and their physical properties i.e., length, Young’s modulus, and the area is used. Further, the final stress is calculated and at last the required stress i.e., forces per unit area is calculated.
Formula used:
$k = \dfrac{{yA}}{l}$
${k_{eq}} = \dfrac{{{k_1}{k_2}}}{{{k_1} + {k_2}}}$
$F = {k_{eq}}\left( x \right)$

Complete step-by-step solution:
As we know stress is defined as force per unit area within the material that arises from outside applied forces.
Given, the young’s modulus for steel say ${y_1} = 120 \times {10^9}N/{m^2}$ and,
The young’s modulus for brass say ${y_2} = 60 \times {10^9}N/{m^2}$
Here, we use the relation between the stress, young’s modulus, cross-section area, and length of the respective wires, which is given by:
${k_1} = \dfrac{{{y_1}{A_1}}}{{{l_1}}} = \dfrac{{120 \times {{10}^9} \times A}}{1}$
Similarly,
${k_2} = \dfrac{{{y_2}{A_2}}}{{{l_2}}} = \dfrac{{60 \times {{10}^9} \times A}}{1}$
We have to calculate the ${k_{eq}}$
${k_{eq}} = \dfrac{{{k_1}{k_2}}}{{{k_1} + {k_2}}}$
Substitute the values of ${k_1}$ and${k_2}$ in the above equation,
${k_{eq}} = \dfrac{{120 \times 60 \times {{10}^9} \times A}}{{180}}$
$\Rightarrow {k_{eq}} = 40 \times {10^9} \times A$
Since we know that
$F = {k_{eq}}\left( x \right)$
where $x$ is the net elongation produced in the wire.
On putting the value of ${k_{eq}}$ and $x$in the above equation, we get;
$F = \left( {40 \times {{10}^9}} \right)A.\left( {0.2 \times {{10}^{ - 3}}} \right)$
$\therefore \dfrac{F}{A} = 8 \times {10^6}N/{m^2}$
Since, to determine the stress the formula required is,
Stress = $\dfrac{F}{A}$
Therefore,
Stress = $8 \times {10^6}N/{m^2}$
Therefore, the stress experienced by the two, brass and steel wires of length $1\;m$ each with areas of cross-section and are connected in series, and one end of the combined wire is connected to a rigid support, and the other end is subjected to elongation is given by the above equation.
As no option is matching.

Additional Information: The maximum stress that a material can experience before it breaks is called the breaking stress or ultimate tensile stress. The word tensile means the material is under tension.

Note: There are different types of stress but the basic concept is the same i.e. stress is defined as force per unit area. Stress experienced by an element or an object depends on the area, length, and also young modulus of the element or object.