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Question: When a wire of length 10cm is subjected to a force of 100N along its length, the lateral strain prod...

When a wire of length 10cm is subjected to a force of 100N along its length, the lateral strain produced is 0.01×10−30.01 \times {10^{ - 3}}. The Poisson’s ratio was found to be 0.4. If the area of cross-section of wire is 0.025m20.025{m^2}, its Young’s modulus is:
(A).1.6×108N/m2 (B).2.5×1010N/m2 (C).12.5×1011N/m2 (D).16×1010N/m2  (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} \\\

Explanation

Solution

Hint- In order to solve this question, firstly we will find the longitudinal stress using the formula of poisson ratio i.e. poisson′s ratio = lateralstrainlongitudinal strainpoisson'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=Normal stresslongitudinal strainY = \dfrac{\text{Normal}{\text{ stress}}}{\text{longitudinal}{\text{ strain}}} to get the required answer.

Complete step-by-step solution -
Formula used-
1. poisson’s ratio = lateral strainlongitudinal strain\text{poisson's} {\text{ ratio = }}\dfrac{{lateral{\text{ strain}}}}{{longitudinal{\text{ strain}}}}
2. Y=Normal stresslongitudinal strainY = \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 = 0.025m20.025{m^2}
And Lateral strain =0.01×10−3 = 0.01 \times {10^{ - 3}}
Firstly, we will apply the formula of poisson’s ratio-
poisson′s ratio = lateral strainlongitudinal strainpoisson's{\text{ ratio = }}\dfrac{{lateral{\text{ strain}}}}{{longitudinal{\text{ strain}}}}
Or longitudinal stress = lateral strainPoisson′s ratiolongitudinal{\text{ stress = }}\dfrac{{lateral{\text{ strain}}}}{{Poisson's{\text{ ratio}}}}
Now substituting the values of Poisson’s ratio = 0.4 and Lateral strain is 0.01×10−30.01 \times {10^{ - 3}},
We get-
=0.01×10−30.4..........(1)= \dfrac{{0.01 \times {{10}^{ - 3}}}}{{0.4}}..........\left( 1 \right)
Now we will apply the young’s modulus,
Y=Normal stresslongitudinal strainY = \dfrac{{Normal{\text{ stress}}}}{{longitudinal{\text{ strain}}}}
Where (Normal stress = ForceArea)\left( {Normal{\text{ stress = }}\dfrac{{Force}}{{Area}}} \right) and substituting the value of equation 1,
We get-
Y=FA×(0.01×10−30.4)Y = \dfrac{F}{{A \times \left( {\dfrac{{0.01 \times {{10}^{ - 3}}}}{{0.4}}} \right)}}
Y=100×0.40.025×0.01×10−3Nm−2Y = \dfrac{{100 \times 0.4}}{{0.025 \times 0.01 \times {{10}^{ - 3}}}}N{m^{ - 2}}
Y=1.6×108Nm−2Y = 1.6 \times {10^8}N{m^{ - 2}}
Therefore, we conclude that if the area of cross-section of wire is 0.025m20.025{m^2}, then its Young’s modulus will be Y=1.6×108Nm−2Y = 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).