Why is the maximum height of a mountain about $\sim 10 \ km$ on Earth?

(N/A) The maximum height of a mountain on Earth depends upon the shear modulus of the rock.
$A$ mountain base is not under uniform compression,and this provides some shearing stress to the rocks,under which they can flow.
The stress due to all the material on the top should be less than the critical shearing stress at which the rocks flow.
At the bottom of a mountain of height $h$,the force per unit area due to the weight of the mountain is $h \rho g$,where $\rho$ is the density of the material of the mountain and $g$ is the acceleration due to gravity.
The material at the bottom experiences this force in the vertical direction,and the sides of the mountain are free.
This creates a shear component,which is approximately $h \rho g$ itself.
The elastic limit for a typical rock is $30 \times 10^{7} \ N \ m^{-2}$.
Therefore,$h \rho g = \text{elastic limit}$.
$h \rho g = 30 \times 10^{7}$.
$h = \frac{30 \times 10^{7}}{\rho g}$.
Substituting $\rho = 3 \times 10^{3} \ kg \ m^{-3}$ and $g = 10 \ m \ s^{-2}$:
$h = \frac{30 \times 10^{7}}{3 \times 10^{3} \times 10} = 10^{4} \ m = 10 \ km$.
This value is consistent with the height of Mt. Everest $(8848 \ m)$.