Question

Assuming that the shear stress at the base of a mountain is equal to the force per unit area to its weight, calculate the maximum possible height of a mountain on the earth if the breaking stress of a typical rock is $30 \times {10^7}{\text{N}}{{\text{m}}^{ - 2}}$ and specific gravity is $3 \times {10^3}{\text{kg}}{{\text{m}}^{ - 3}}$.
A) $10{\text{km}}$
B) ${\text{8km}}$
C) ${\text{7km}}$
D) ${\text{6km}}$

Answer 1

The force acting at the base of the mountain is the force due to gravity or its weight and the shear stress at the base is mentioned to be due to this force. The shear stress must be less than the given breaking stress of a rock. So this inequality will help us to calculate the maximum height of the mountain.

Formulas used:
The weight of a body is given by, $W = \rho Vg$ where $\rho$ is the density of the body, $V$ is the volume of the body and $g$ is the acceleration due to gravity.
The shear stress acting on a body is given by, ${\sigma _s} = \dfrac{F}{A}$ where $F$ is the force acting on the body and the $A$ is the area of the body.

Complete step by step answer:
Step 1: List the parameters given in the question.
The breaking stress of rock is given to be ${\sigma _{breaking}} = 30 \times {10^7}{\text{N}}{{\text{m}}^{ - 2}}$.
The specific gravity is given to be $\rho = 3 \times {10^3}{\text{kg}}{{\text{m}}^{ - 3}}$ .
We have to find the maximum height $h$ of the mountain.
Step 2: Express the relation for the shearing stress at the base of the mountain.
The shear stress at the base of the mountain can be expressed as ${\sigma _s} = \dfrac{W}{A}$ ---------- (1) where $W$ is the weight or the force due to gravity and $A$ is the area of the mountain.
We express the weight of the mountain as
$W = \rho Vg$ -------- (2) where $\rho$ is the specific gravity, $V$ is the volume of the mountain and $g$ is the acceleration due to gravity.
Substituting equation (2) in (1) we get, ${\sigma _s} = \dfrac{{\rho Vg}}{A}$.
Since the volume can be expressed as $V = Ah$, the above equation can also be expressed as ${\sigma _s} = \dfrac{{\rho Ahg}}{A} = \rho hg$.
Thus we have the shearing stress at the base as ${\sigma _s} = \rho hg$.
Step 3: Apply the condition for the mountain to remain intact to find the maximum height.
For the mountain to restrain from breaking or crumbling, the shear stress at the base of the mountain must be less than the breaking stress of rock.
i.e., ${\sigma _s} \leqslant {\sigma _{breaking}}$ or $\rho hg \leqslant {\sigma _{breaking}}$ --------- (3)
Now substituting for $\rho = 3 \times {10^3}{\text{kg}}{{\text{m}}^{ - 3}}$, $g = 10{\text{m}}{{\text{s}}^{ - 2}}$ and ${\sigma _{breaking}} = 30 \times {10^7}{\text{N}}{{\text{m}}^{ - 2}}$ in equation (3) we get,
$\Rightarrow 3 \times {10^3} \times h \times 10 \leqslant 30 \times {10^7}$
$\Rightarrow h = {10^4}{\text{m}}$
Thus we have the maximum height of the mountain to be $10{\text{km}}$.

Hence the correct option is A.

Note:
Here we assumed the acceleration due to gravity to be $g = 10{\text{m}}{{\text{s}}^{ - 2}}$. The given specific gravity expresses the density of the mountain. The original formula for weight is $W = mg$ but as the mass of a body can be expressed in terms of its density and volume as $m = \rho V$, we replace the mass $m$ by this relation and obtain $W = \rho Vg$.