
Possible value of Poisson's ratio is
\[(A)1\]
\[(B)0.9\]
\[(C)0.8\]
\[(D)0.4\]
Answer
501.9k+ views
Hint: Poisson’s ratio has no unit due to one quantity divided by another quantity with the identical unit, so there’s no unit for Poisson’s ratio. Poisson's ratio is defined as the ratio of width per unit to change in length per unit. Poisson's ratio is transverse lateral strain to longitudinal strain within the direction of the stretching force.
Complete step-by-step solution:
The Poisson’s ratio of stable and linear elastic material will be greater than \[1\] or less than\[0.5\].
Let \[Y,K,n\] and \[\sigma \] are the young's modulus, bulk modulus, modulus of rigidity, and Poisson’s ratio respectively,
Therefore, \[Y = 3K\left( {1 - 2\sigma } \right)\] it is a standard formula
\[Y = 2n\left( {1 + \sigma } \right)\] this also standard formula
Hence we get, \[3K\left( {1 - 2\sigma } \right) = 2n\left( {1 + \sigma } \right)\]
Now we are assuming that \[K,n\] are always positive so that we can define in two cases,
Case(i): if \[\sigma \]is positive, then the right-hand side is always positive. So, the left-hand side must be positive, therefore\[2\sigma < 1\]or \[\sigma < 0.5\]
Case (ii): if \[\sigma \] is negative, then the left-hand side will always be positive, therefore \[1 + \sigma > 0\]or \[\sigma > - 1\]
Thus the intensity volume of Poisson’s ratio lies between \[ - 1 < \sigma < 0.5\] from the requirement of Young's modulus. Most of the materials have Poisson’s ratio value ranging between \[0\]to\[0.5\].
From the given options, the only possible value is\[0.4\] .
Hence, option D is correct.
Note:Most materials are having Poisson's ratio values ranging between \[0\] and \[0.5\]. A rubber having a ratio near\[0.5\], it is an Incompressible material. This ratio is called after the French physicist and great mathematician Siméon Poisson. Poisson's ratios exceeding \[0.5\] are allowed in anisotropic materials. And also hexagonal honeycombs are able to exhibit Poisson's ratio of\[1\], and if they need arranged hexagonal cells in certain directions which is greater than\[1\].
Complete step-by-step solution:
The Poisson’s ratio of stable and linear elastic material will be greater than \[1\] or less than\[0.5\].
Let \[Y,K,n\] and \[\sigma \] are the young's modulus, bulk modulus, modulus of rigidity, and Poisson’s ratio respectively,
Therefore, \[Y = 3K\left( {1 - 2\sigma } \right)\] it is a standard formula
\[Y = 2n\left( {1 + \sigma } \right)\] this also standard formula
Hence we get, \[3K\left( {1 - 2\sigma } \right) = 2n\left( {1 + \sigma } \right)\]
Now we are assuming that \[K,n\] are always positive so that we can define in two cases,
Case(i): if \[\sigma \]is positive, then the right-hand side is always positive. So, the left-hand side must be positive, therefore\[2\sigma < 1\]or \[\sigma < 0.5\]
Case (ii): if \[\sigma \] is negative, then the left-hand side will always be positive, therefore \[1 + \sigma > 0\]or \[\sigma > - 1\]
Thus the intensity volume of Poisson’s ratio lies between \[ - 1 < \sigma < 0.5\] from the requirement of Young's modulus. Most of the materials have Poisson’s ratio value ranging between \[0\]to\[0.5\].
From the given options, the only possible value is\[0.4\] .
Hence, option D is correct.
Note:Most materials are having Poisson's ratio values ranging between \[0\] and \[0.5\]. A rubber having a ratio near\[0.5\], it is an Incompressible material. This ratio is called after the French physicist and great mathematician Siméon Poisson. Poisson's ratios exceeding \[0.5\] are allowed in anisotropic materials. And also hexagonal honeycombs are able to exhibit Poisson's ratio of\[1\], and if they need arranged hexagonal cells in certain directions which is greater than\[1\].
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