Question

What is the relation between Y, K and $\eta$?
(A) ${\displaystyle \frac{3}{Y}}={\displaystyle \frac{1}{3K}}+{\displaystyle \frac{1}{\eta }}$
(B) ${\displaystyle \frac{1}{Y}}={\displaystyle \frac{1}{9K}}+{\displaystyle \frac{1}{\eta }}$
(C) ${\displaystyle \frac{9}{Y}}={\displaystyle \frac{1}{K}}+{\displaystyle \frac{3}{\eta }}$
(D) ${\displaystyle \frac{1}{\eta }}={\displaystyle \frac{1}{K}}+{\displaystyle \frac{1}{Y}}$

Answer 1

Hint: Young’s modulus Y is the ratio of the tensile stress to strain, while the bulk modulus K is the ratio of the volumetric stress to volumetric strain. These quantities have specific relations with each other and the modulus of rigidity .

Complete step by step answer:
The Young’s modulus, bulk modulus and modulus of elasticity are all mechanical properties of a material. In this question we are asked to find a relation between the following:
-Young’s modulus Y
-Bulk modulus K
-Modulus of rigidity $\eta$
We know that the relationship between these quantities is given by:
$\Rightarrow Y=3K(1-2\sigma )$ [$\sigma$ is the Poisson’s ratio]
We now try to write this relation in terms of $\sigma$ on the LHS. So cross multiplication will give us:
$\Rightarrow {\displaystyle \frac{Y}{3K}}=1-2\sigma$
$\Rightarrow 2\sigma =1-{\displaystyle \frac{Y}{3K}}$
$\Rightarrow \sigma ={\displaystyle \frac{1}{2}}\left(1-{\displaystyle \frac{Y}{3K}}\right)$ [Let this be Eq. 1]
We have another relation between the Young’s modulus and modulus of rigidity as:
$\Rightarrow Y=2\eta (1+\sigma )$
Again, solving for $\sigma$ gives us:
$\Rightarrow {\displaystyle \frac{Y}{2\eta }}=1+\sigma$
$\Rightarrow \sigma ={\displaystyle \frac{Y}{2\eta }}-1$ [Let this be Eq. 2]
As the LHS of both Eq. 1 and Eq. 2 are same, we equate the two to get:
$\Rightarrow {\displaystyle \frac{1}{2}}\left(1-{\displaystyle \frac{Y}{3K}}\right)={\displaystyle \frac{Y}{2\eta }}-1$
Simplifying this equation brings us to:
$\Rightarrow 1-{\displaystyle \frac{Y}{3K}}={\displaystyle \frac{Y}{\eta }}-2$
$\Rightarrow 1+2={\displaystyle \frac{Y}{\eta }}+{\displaystyle \frac{Y}{3K}}=Y\left({\displaystyle \frac{1}{\eta }}+{\displaystyle \frac{1}{3K}}\right)$
Taking Y on the LHS, gives us the final relation as:
${\displaystyle \frac{3}{Y}}=\left({\displaystyle \frac{1}{\eta }}+{\displaystyle \frac{1}{3K}}\right)$
Hence, the correct answer is option (A).

Note:
Young’s modulus is helpful in predicting the extent of compression or elongation of a body under stress. Similarly, the bulk modulus is helpful in determining the change in volume of an object when it is subjected to some pressure. The modulus of rigidity or the bulk modulus tells about the behaviour of a material when a horizontal force is applied in a direction tangential to its surface or when a material is twisted during torsional testing.