The relation between Y (Young's modulus), K (bulk modulus) and `eta` (shear modulus) is

Options

A. `(9)/(Y) = (1)/(B) + (3)/(eta)`
B. `(1)/(Y) = (1)/(3 eta) + (1)/(9B)`
C. `(9)/(Y) = (1)/(eta) + (3)/(B)`
D. `(1)/(eta) = (1)/(B) + (1)/(Y)`

Explanation

To find the relation between Young's modulus (Y), bulk modulus (K), and shear modulus (η), we can start with the known formulas for these moduli in terms of Poisson's ratio (σ).

### Step-by-Step Solution:

1. **Write the formulas for Young's modulus (Y) in terms of K and σ:**
\[ Y = 3K(1 - 2\sigma) \]

2. **Write the formula for Young's modulus (Y) in terms of η and σ:**
\[ Y = 2η(1 + \sigma) \]

3. **From the first equation, express σ in terms of Y and K:**
\[ Y = 3K(1 - 2\sigma) \implies \frac{Y}{3K} = 1 - 2\sigma \implies 2\sigma = 1 - \frac{Y}{3K} \]
\[ \sigma = \frac{1}{2} - \frac{Y}{6K} \]

4. **Substitute this expression for σ into the second equation:**
\[ Y = 2η\left(1 + \left(\frac{1}{2} - \frac{Y}{6K}\right)\right) \]
\[ Y = 2η\left(\frac{3}{2} - \frac{Y}{6K}\right) \]

5. **Distribute 2η:**
\[ Y = 3η - \frac{Yη}{3K} \]

6. **Rearranging gives us:**
\[ Y + \frac{Yη}{3K} = 3η \]

7. **Factor out Y:**
\[ Y\left(1 + \frac{η}{3K}\right) = 3η \]

8. **Finally, solve for the relationship:**
\[ Y = \frac{3η}{1 + \frac{η}{3K}} \implies \frac{1}{Y} = \frac{1}{3η} + \frac{1}{3K} \]

Thus, the relation between Young's modulus (Y), bulk modulus (K), and shear modulus (η) is:
\[ \frac{1}{Y} = \frac{1}{3η} + \frac{1}{3K} \]