91Ó°ÊÓ

The thickness of the mountain is \(d = 2\;{\rm{km}}\).

In this problem, the pressure on the mountain can be calculated by dividing the weight of the ice (mg) with the area (A).

The relation ofpressureis given by,

\(\begin{array}{l}P = \frac{F}{A}\\P = \frac{{mg}}{{\left( {\frac{V}{d}} \right)}}\end{array}\)…… (i)

Here, \(F\) is the weight of the ice, \(m\) is the mass of the ice, \(g\) is the gravitational acceleration and \(V\)is the volume.

The relation of volume is given by,

\(V = \frac{m}{\rho }\)

Here, \(\rho \) is the density of the ice.

On plugging the values in the equation (i),

\(\begin{array}{l}P = \frac{{mg}}{{\left( {\frac{m}{{\rho d}}} \right)}}\\P = \rho gd\\P = \left( {917\;{\rm{kg/}}{{\rm{m}}^3}} \right)\left( {9.80\;{\rm{m/}}{{\rm{s}}^2}} \right)\left( {2\;{\rm{km}} \times \frac{{1000\;{\rm{m}}}}{{1\;{\rm{km}}}}} \right)\\P = 1.79 \times {10^7}\;{\rm{Pa}}\end{array}\)

Thus, the pressure on the mountains is \(1.79 \times {10^7}\;{\rm{Pa}}\).