Question

The proportional limit of steel is $8 \times {10^8}N/{m^2}$ and its Young’s modulus is $2 \times {10^{11}}N/{m^2}$ . The maximum elongation, a one metre long steel wire can be given without exceeding the proportional limit is
(A) $2mm$
(B) $4mm$
(C) $1mm$
(D) $8mm$

Answer 1

Hint
As we already know that, stress is directly proportional to strain
$\therefore $ stress $\infty $strain
So, stress = $\gamma \times $ strain (where, $\gamma $ is the Young’s Modulus)
Then, we will use the expression $\dfrac{{\Delta L}}{L} = $ strain (where, $\Delta L$ is the stress and $L$ is the length of wire)

Complete step by step solution
The force which is applied to a material is called Stress. It is denoted by $\sigma $. The formula for calculating stress is-
$\sigma = \dfrac{F}{A}$ (where, $F$ is the force and $A$ is the area of cross-section)
Strain is the deformation or displacement of material that results from an applied stress. It is denoted by $\varepsilon $. The formula for calculating strain is-
$\varepsilon = \dfrac{{L - {L_0}}}{{{L_0}}}
 \varepsilon = \dfrac{{\Delta L}}{{{L_0}}} \cdots (1) $
Where, $L$ is the length after load is applied and ${L_0}$ is the original length.
According to question, it is given that-
Proportional limit of steel = stress = $8 \times {10^8}N/{m^2}$
Young’s Modulus, $\gamma = $ $2 \times {10^{11}}N/{m^2}$
Length of wire, $L = 1m$
We know that, stress is directly proportional to strain
stress $\infty $strain
$\therefore \sigma = \gamma \times \varepsilon \cdots (2)$
Putting the values of stress and Young’s Modulus in the equation $(2)$
$8 \times {10^8} = 2 \times {10^{11}} \times \varepsilon
 \Rightarrow \varepsilon = \dfrac{{8 \times {{10}^8}}}{{2 \times {{10}^{11}}}} $
From the above calculation we get the value of strain-
$\varepsilon = 4 \times {10^{ - 3}}$
Now, using equation $(1)$
$\varepsilon = \dfrac{{\Delta L}}{{{L_0}}}$
Putting the values of strain and length of wire in the above equation, we get
$\dfrac{{\Delta L}}{1} = 4 \times {10^{ - 3}}$
So, $\Delta L = 4 \times {10^{ - 3}}
   \therefore \Delta L = 4mm $
Therefore, the correct answer for this question is option (B).

Note
The most widely used and common method to establish the relationship between stress and strain is the stress-strain curve for a particular material. The stress-strain curve which takes instantaneous value of cross-sectional area and length for stress and strain is called a true stress-strain diagram.