A wire is subjected to a longitudinal strain of 0.05. If its material has a Poisson’s ratio 0.25, the lateral strain experienced by it is
A. 0.00625
B. 0.125
C. 0.0125
D. 0.0625
Answer
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Hint: Young’s modulus is the ratio of longitudinal stress over strain, and poisson’s ratio is the ratio of lateral increment or decrement of the material. To find the solution of the given question, write down all the given physical quantities and apply the formula of Poisson’s ratio to find the lateral strain experienced by wire.
Formula used:
$\sigma = \dfrac{{{\text{Lateral constraint}}}}{{{\text{Longitudinal strain}}}}$
Complete answer:
The strain is defined as the change in dimension divided by the original dimension. Poisson’s ratio is defined as the ratio of transverse contraction strain to the longitudinal extension strain in the direction of the stretching force.
Mathematically, it can be written as,
$\sigma = \dfrac{{{\text{Lateral constraint}}}}{{{\text{Longitudinal strain}}}}$
$ \Rightarrow {\text{Lateral conststain}} = \sigma \times {\text{Longitudinal strain}}$
$ \Rightarrow {\text{Lateral strain}} = 0.25 \times 0.05 = 0.0125$
So, the correct answer is “Option c”.
Additional Information:
Poisson’s ratio can also be explained through an example such as when a rubber band is stretched it tends to become thinner. Poisson’s ratio increases with time and the temperature. It remains approximately constant within elastic limits for a material. When an external force is applied to an object it undergoes deformation. If the direction of the force acting on the body is parallel to the plane of the object, the deformation occurs along the plane and this stress experienced by the object is known as shear stress.
Note:
Poisson’s ratio is defined as, when a material is stretched in one direction, it tends to compress in the direction perpendicular to that of force application and vice versa, the measure of this phenomenon is known as Poisson’s ratio. Poisson’s ratio is positive for a tensile deformation, and is negative for a compressive deformation. The value of Poisson’s ratio lies in the range 0 to 0.5 for most of the materials.
Formula used:
$\sigma = \dfrac{{{\text{Lateral constraint}}}}{{{\text{Longitudinal strain}}}}$
Complete answer:
The strain is defined as the change in dimension divided by the original dimension. Poisson’s ratio is defined as the ratio of transverse contraction strain to the longitudinal extension strain in the direction of the stretching force.
Mathematically, it can be written as,
$\sigma = \dfrac{{{\text{Lateral constraint}}}}{{{\text{Longitudinal strain}}}}$
$ \Rightarrow {\text{Lateral conststain}} = \sigma \times {\text{Longitudinal strain}}$
$ \Rightarrow {\text{Lateral strain}} = 0.25 \times 0.05 = 0.0125$
So, the correct answer is “Option c”.
Additional Information:
Poisson’s ratio can also be explained through an example such as when a rubber band is stretched it tends to become thinner. Poisson’s ratio increases with time and the temperature. It remains approximately constant within elastic limits for a material. When an external force is applied to an object it undergoes deformation. If the direction of the force acting on the body is parallel to the plane of the object, the deformation occurs along the plane and this stress experienced by the object is known as shear stress.
Note:
Poisson’s ratio is defined as, when a material is stretched in one direction, it tends to compress in the direction perpendicular to that of force application and vice versa, the measure of this phenomenon is known as Poisson’s ratio. Poisson’s ratio is positive for a tensile deformation, and is negative for a compressive deformation. The value of Poisson’s ratio lies in the range 0 to 0.5 for most of the materials.
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