The relationship between Young’s modulus Y, bulk modulus K and modulus of rigidity η is

To find the correct relationship between Young’s modulus ($Y$), bulk modulus ($K$), and modulus of rigidity ($\eta$), we can use the standard relations connecting these elastic constants with Poisson's ratio ($\sigma$).

The first standard relation connects Young's modulus, bulk modulus, and Poisson's ratio:
$Y=3K(1-2\sigma )$

From this, we can express $1-2\sigma$ as:
$1-2\sigma =\frac{Y}{3K}$      --- (Equation 1)

The second standard relation connects Young's modulus, modulus of rigidity, and Poisson's ratio:
$Y=2\eta (1+\sigma )$

From this, we can express $1+\sigma$ as:
$1+\sigma =\frac{Y}{2\eta }$      --- (Equation 2)

To eliminate Poisson's ratio ($\sigma$), we multiply Equation 2 by 2:
$2+2\sigma =\frac{2Y}{2\eta }=\frac{Y}{\eta }$      --- (Equation 3)

Now, we add Equation 1 and Equation 3 together:
$(1-2\sigma )+(2+2\sigma )=\frac{Y}{3K}+\frac{Y}{\eta }$

Simplifying the left-hand side:
$3=Y\left(\frac{1}{3K}+\frac{1}{\eta }\right)$

Taking the common denominator on the right-hand side:
$3=Y\left(\frac{\eta +3K}{3\eta K}\right)$

Rearranging the equation to solve for Young's modulus ($Y$):
$Y=\frac{9\eta K}{\eta +3K}$

Therefore, the correct relationship matches the first option.