Question

The Young’s modulus of a material is $2 \times {10^{11}}N/{m^2}$ and its elastic limit is $1.8 \times {10^8}N/{m^2}.$ For a wire of 1m length of this material, the maximum elongation achievable is
(A). 0.2mm
(B). 0.3mm
(C). 0.4mm
(D). 0.9mm

Answer 1

- Hint: In order to solve this question, we will use the concept of stress and strain. As we know that when a stress is applied on a body and then the stress is released after sometime the body restores its original dimension if the stress applied is in limit and if we cross the limit, the body will be in permanent deformation.

Complete step-by-step solution -
Elastic limit:- This limit of strain or corresponding stress is called elastic limit of the material .
For solving the problem we will use the Hooke's law which says that stress is directly proportional to the strain up to elastic limit.
i.e.
stress $\alpha$ strain
$\Rightarrow \dfrac{F}{A}\alpha \dfrac{{\Delta l}}{l} \\\ \Rightarrow \dfrac{F}{A} = \dfrac{{Y\Delta l}}{l} \\\$
Where
F is the force
A is the area
$l$ is the length
Y is the young’s modulus constant
$\Delta l$ is the change in the length.
Given that
Y = $2 \times {10^{11}}N/{m^2}$
As we know that maximum stress is equal to the elastic limit.
Elastic limit = $1.8 \times {10^8}N/{m^2}.$
The change in length can be calculated using the above formula
$\Rightarrow \dfrac{F}{A} = \dfrac{{Y\Delta l}}{l}$
Substituting the values in the above formula
$\Rightarrow \dfrac{F}{A} = \dfrac{{Y\Delta l}}{l} = 1.8 \times {10^8}N/{m^2} \\\ \Rightarrow \dfrac{{\Delta l}}{1} = \dfrac{{1.8 \times {{10}^8}N/{m^2}}}{{2 \times {{10}^{11}}N/m}} \\\ \Rightarrow \Delta l = 0.9mm \\\$
Hence correct answer is 0.9mm
Option D is the correct answer.

Note- Young's modulus is the property of material that defines the capacity of material to withstand the change in length when applied to some tension or compression. Higher the young’s modulus more will be the withstanding capacity.