Question

When a wire of length 10cm is subjected to a force of 100N along its length, the lateral strain produced is $0.01 \times {10^{ - 3}}$. The Poisson’s ratio was found to be 0.4. If the area of cross-section of wire is $0.025{m^2}$, its Young’s modulus is:
$
  (A). 1.6 \times {10^8}N/{m^2} \\
  (B). 2.5 \times {10^{10}}N/{m^2} \\
  (C). 12.5 \times {10^{11}}N/{m^2} \\
  (D). 16 \times {10^{10}}N/{m^2} \\
 $

Answer 1

Hint- In order to solve this question, firstly we will find the longitudinal stress using the formula of poisson ratio i.e. $poisson's{\text{ ratio = }}\dfrac{\text{lateral} {\text{strain}}} {\text{longitudinal} {\text{ strain}}}$. Then we will use the formula of young’s modulus i.e. $Y = \dfrac{\text{Normal}{\text{ stress}}}{\text{longitudinal}{\text{ strain}}}$ to get the required answer.

Complete step-by-step solution -
Formula used-
1. $\text{poisson's} {\text{ ratio = }}\dfrac{{lateral{\text{ strain}}}}{{longitudinal{\text{ strain}}}}$
2. $Y = \dfrac{{Normal{\text{ stress}}}}{{longitudinal{\text{ strain}}}}$
Here we are given that-
Length of wire = 10cm
Poisson’s ratio = 0.4
Area of cross-section of wire = $0.025{m^2}$
And Lateral strain $ = 0.01 \times {10^{ - 3}}$
Firstly, we will apply the formula of poisson’s ratio-
$poisson's{\text{ ratio = }}\dfrac{{lateral{\text{ strain}}}}{{longitudinal{\text{ strain}}}}$
Or $longitudinal{\text{ stress = }}\dfrac{{lateral{\text{ strain}}}}{{Poisson's{\text{ ratio}}}}$
Now substituting the values of Poisson’s ratio = 0.4 and Lateral strain is $0.01 \times {10^{ - 3}}$,
We get-
$ = \dfrac{{0.01 \times {{10}^{ - 3}}}}{{0.4}}..........\left( 1 \right)$
Now we will apply the young’s modulus,
$Y = \dfrac{{Normal{\text{ stress}}}}{{longitudinal{\text{ strain}}}}$
Where $\left( {Normal{\text{ stress = }}\dfrac{{Force}}{{Area}}} \right)$ and substituting the value of equation 1,
We get-
$Y = \dfrac{F}{{A \times \left( {\dfrac{{0.01 \times {{10}^{ - 3}}}}{{0.4}}} \right)}}$
$Y = \dfrac{{100 \times 0.4}}{{0.025 \times 0.01 \times {{10}^{ - 3}}}}N{m^{ - 2}}$
$Y = 1.6 \times {10^8}N{m^{ - 2}}$
Therefore, we conclude that if the area of cross-section of wire is $0.025{m^2}$, then its Young’s modulus will be $Y = 1.6 \times {10^8}N{m^{ - 2}}$.

Note- While solving this question, we must know the concept of young’s modulus i.e. it shows the relationship between stress (force per unit area) and strain ( proportional deformation in object). Also one should not forget to put the SI unit along with the answer (here young’s modulus).