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Question: The formula that relates all three elastic constants is. A. \(9/y = 3/n - 1/B\) B. \(9/y = 3/n +...

The formula that relates all three elastic constants is.
A. 9/y=3/n1/B9/y = 3/n - 1/B
B. 9/y=3/n+1/B9/y = 3/n + 1/B
C. 9/y=3/n+2/B9/y = 3/n + 2/B
D. 9/y=3/n2/B9/y = 3/n - 2/B

Explanation

Solution

To build the relation between the three elastic constants, we can use the relations among two elastic constants with each other. To determine the answer in the form of given options, we can do some simple rearrangement and calculation with the relations of elastic constants.

Complete step by step solution:
Let yy is the modulus of elasticity, nn is the modulus of rigidity and BB is the bulk modulus.
The relation between yy and nn is,
y=2n(1+μ)y = 2n\left( {1 + \mu } \right) (1)

Here μ\mu is the Poisson’s ratio.

The relation between yy and BB is,
y=3B(12μ)y = 3B\left( {1 - 2\mu } \right) (2)

From equation (2), the expression of the Poisson’s ratio is,
y=3B(12μ) y3B1=2μ μ=12y6B\begin{array}{l} y = 3B\left( {1 - 2\mu } \right)\\\ \dfrac{y}{{3B}} - 1 = - 2\mu \\\ \mu = \dfrac{1}{2} - \dfrac{y}{{6B}} \end{array} (3)

Now from equation (1) and (3), the relation between all three elastic constant is,

y = 2n\left( {1 + \left( {\dfrac{1}{2} - \dfrac{y}{{6B}}} \right)} \right)\\\ y = 2n + n - \dfrac{{ny}}{{3B}}\\\ y = \dfrac{{9nB - ny}}{{3B}}\\\ 3yB + nY = 9nB \end{array}$$ On further solving the above equation $$\begin{array}{l} y = \dfrac{{9nB}}{{3B + n}}\\\ \dfrac{9}{y} = \dfrac{3}{n} + \dfrac{1}{B} \end{array}$$ **Therefore, the option (B) is the correct answer that is $9/y = 3/n + 1/B$.** **Note:** The relation of elastic constants consist of the value of Poisson's ratio, so do not forget to remove the value of Poisson's ratio from the relations. To remove Poisson's ratio, replace the value of Poisson's ratio arrived from equation (2) in equation (1).