
For a material $Y = 6.6 \times {10^{10}}N{m^{ - 2}}$ and bulk modulus $K = 11 \times {10^{10}}N{m^{ - 2}}$, then its Poisson’s ratio is
A) $0.8$
B) $0.35$
C) $0.7$
D) $0.4$
Answer
564k+ views
Hint:Recall that $Y$ is the Young’s modulus and $K$ is the bulk modulus. To find the Poisson’s ratio, we shall use the relation that relates all these values together.
Formulas Used:
Young’s Modulus, $Y = 3B(1 - 2\sigma )$
Where, $B$ is the bulk modulus and $\sigma $ is the Poisson’s ratio.
Complete step by step answer:
First, we check if all the given values are in SI units. Then we use the formula for Young modulus to relate Young’s Modulus, Bulk modulus, and Poisson’s ratio in a single formula.
The formula is, $Y = 3B(1 - 2\sigma )$
Where, $Y$ is the Young’s Modulus, $B$ is the bulk modulus and $\sigma $ is the Poisson’s ratio.
We will put the given values in this formula accordingly. We get, $6.6 \times {10^{10}} = 3 \times 11 \times {10^{10}}(1 - 2\sigma )$
On further solving this, we get $(1 - 2\sigma ) = \dfrac{{6.6 \times {{10}^{10}}}}{{3 \times 11 \times {{10}^{{{10}^{}}}}}}$
Which gives, $(1 - 2\sigma ) = \dfrac{{6.6}}{{33}}$ $ \Rightarrow 1 - 2\sigma = 0.2$
On transposing values as required we get $2\sigma = 1 - 0.2$
This gives $\sigma = \dfrac{{0.8}}{2}$
$ \Rightarrow \sigma = 0.4$
This is the required answer. Poisson’s ratio for given values in $0.4$
Additional Information: Poisson's ratio is the ratio of transverse strain to corresponding axial strain on a material stressed along one axis. static Poisson's ratio can be determined by measuring the axial and lateral deformations of the uniaxial compression test in the rock sample. Static Poisson's ratio in a rock depends on lithology, confining stress, pore pressure, and porosity of the rock. Laboratory test results show that static Poisson's ratio increases with increase in porosity.
Note:Always check if the values given are in Si units or not. If not, convert them before using them in the question. The formula used above is of great importance as it relates all the three values mentioned in the question i.e. Young’s Modulus, Bulk Modulus, and Poisson’s ratio.
Formulas Used:
Young’s Modulus, $Y = 3B(1 - 2\sigma )$
Where, $B$ is the bulk modulus and $\sigma $ is the Poisson’s ratio.
Complete step by step answer:
First, we check if all the given values are in SI units. Then we use the formula for Young modulus to relate Young’s Modulus, Bulk modulus, and Poisson’s ratio in a single formula.
The formula is, $Y = 3B(1 - 2\sigma )$
Where, $Y$ is the Young’s Modulus, $B$ is the bulk modulus and $\sigma $ is the Poisson’s ratio.
We will put the given values in this formula accordingly. We get, $6.6 \times {10^{10}} = 3 \times 11 \times {10^{10}}(1 - 2\sigma )$
On further solving this, we get $(1 - 2\sigma ) = \dfrac{{6.6 \times {{10}^{10}}}}{{3 \times 11 \times {{10}^{{{10}^{}}}}}}$
Which gives, $(1 - 2\sigma ) = \dfrac{{6.6}}{{33}}$ $ \Rightarrow 1 - 2\sigma = 0.2$
On transposing values as required we get $2\sigma = 1 - 0.2$
This gives $\sigma = \dfrac{{0.8}}{2}$
$ \Rightarrow \sigma = 0.4$
This is the required answer. Poisson’s ratio for given values in $0.4$
Additional Information: Poisson's ratio is the ratio of transverse strain to corresponding axial strain on a material stressed along one axis. static Poisson's ratio can be determined by measuring the axial and lateral deformations of the uniaxial compression test in the rock sample. Static Poisson's ratio in a rock depends on lithology, confining stress, pore pressure, and porosity of the rock. Laboratory test results show that static Poisson's ratio increases with increase in porosity.
Note:Always check if the values given are in Si units or not. If not, convert them before using them in the question. The formula used above is of great importance as it relates all the three values mentioned in the question i.e. Young’s Modulus, Bulk Modulus, and Poisson’s ratio.
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