Question

The relation between young’s modulus $Y$, bulk modulus $K$, and modulus of elasticity $\sigma $ is
A) $\dfrac{1}{y} = \dfrac{1}{k} + \dfrac{3}{\eta }$
B) $\dfrac{3}{y} = \dfrac{1}{\eta } + \dfrac{1}{{3k}}$
C) $\dfrac{1}{y} = \dfrac{3}{\eta } + \dfrac{1}{{3k}}$
D) $\dfrac{1}{\eta } = \dfrac{3}{y} + \dfrac{1}{{3k}}$

Answer 1

Hint: Find the two relations of young’s modulus with bulk modulus and modulus of rigidity respectively. Then try to eliminate the stress applied $\left( \sigma \right)$ from both the relations.

Complete answer:
So we have to interrelate the young’s modulus, bulk modulus, and modulus of elasticity. To introduce young’s modulus $Y$, we have to provide longitudinal strain in the object. Since there is a longitudinal strain, so with that a lateral strain would also be there which helps to introduce the term Poisson’s ratio $\sigma$.
To introduce bulk modulus $K$, we have to provide volumetric strain in all directions of the object.
Let $\alpha $=longitudinal strain per unit stress
And $\beta $=lateral strain per unit stress.
Then Poisson’s ratio, \[\sigma = \dfrac{{\text{lateral strain}}}{{\text{longitudinal strain}}} = \dfrac{\beta }{\alpha }\]
Young’s modulus, $Y = \dfrac{{\text{longitudinal stress}}}{{\text{longitudinal strain}}} = \dfrac{1}{\alpha }$
And Bulk modulus, $K = \dfrac{{\text{hydrostatic strain}}}{{\text{volumetric strain}}} = \dfrac{1}{{3\left( {\alpha - 2\beta } \right)}}$
Now to introduce $\sigma $in the formula of bulk modulus, we do as follows
$K = \dfrac{1}{{3\left( {\alpha - 2\beta } \right)}} = \dfrac{1}{{3\alpha \left( {1 - 2\dfrac{\beta }{\alpha }} \right)}}$
Since $\sigma = \dfrac{\beta }{\alpha }$ and $Y = \dfrac{1}{\alpha }$ , so by substituting these values in above equation, we get
$K = \dfrac{Y}{{3\left( {1 - 2\sigma } \right)}} \cdot \cdot \cdot \cdot \cdot \cdot \left( 1 \right)$
Now looking into options, as we see that the options are given in terms of $Y$, $K$ and modulus of rigidity$\left( \eta \right)$. For this we use the formula
$Y = 2\eta \left( {1 + \sigma } \right) \cdot \cdot \cdot \cdot \cdot \cdot \left( 2 \right)$
Now get the value of $\sigma $from both the equation $\left( 1 \right)$and $\left( 2 \right)$
From equation$\left( 1 \right)$,
$
  1 - 2\sigma = \dfrac{Y}{{3K}} \\
   \to 2\sigma = 1 - \dfrac{Y}{{3K}} \\
 $
\[ \Rightarrow \sigma = \dfrac{1}{2}\left( {1 - \dfrac{Y}{{3K}}} \right) \cdot \cdot \cdot \cdot \cdot \cdot \left( 3 \right)\]
Now from equation $\left( 2 \right)$
$\sigma = \dfrac{Y}{{2\eta }} - 1 \cdot \cdot \cdot \cdot \cdot \cdot \left( 4 \right)$
By comparing equations $\left( 3 \right)$ and$\left( 4 \right)$, we have
$
  \dfrac{1}{2}\left( {1 - \dfrac{Y}{{3K}}} \right) = \dfrac{Y}{{2\eta }} - 1 \\
  \dfrac{3}{2} - \dfrac{Y}{{6K}} = \dfrac{Y}{{2\eta }} \Leftrightarrow 3 - \dfrac{Y}{{3K}} = \dfrac{Y}{\eta } \\
 $
On further solving, we get
$
  3 = \dfrac{Y}{\eta } + \dfrac{Y}{{3K}} \\
   \Rightarrow \dfrac{3}{Y} = \dfrac{1}{\eta } + \dfrac{1}{{3K}} \\
 $

$\therefore$ The relation between young’s modulus $Y$, bulk modulus $K$, and modulus of elasticity $\sigma$ is $\dfrac{3}{Y} = \dfrac{1}{\eta } + \dfrac{1}{{3K}}$. Hence, the correct option is B.

Note:
Don’t confuse it with too many modulus. Basically there are three types of modulus in elasticity: Modulus of elasticity or Young’s modulus, Bulk modulus, and Modulus of rigidity or Shear modulus. While finding relations, look at the options to get the idea to which format we have to find the ultimate relationship.