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Whenever a specific external force acts on a particular area, then the value of pressure due to external force on the area can obtain with the help of force and area. The relation between the pressure and the external pressure is a linear one.

The area of eardrum is\(A = 0.20\;{\rm{c}}{{\rm{m}}^{\rm{2}}}\).

The change in altitudes is \(h = 1250\;{\rm{m}}\).

The expression of the difference in pressure is given by,

\(\Delta P = \rho gh\)

Here, \(\Delta P\) is the pressure difference, \(\rho \) is the density of air \(\left( {\rho = 1.23\;{\rm{kg/}}{{\rm{m}}^{\rm{3}}}} \right)\) and \(g\) is the gravitational acceleration\(g = 9.81\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}\).

Substitute all the known values in the above expression.

\(\begin{array}{c}\Delta P = \left( {1.23\;{\rm{kg/}}{{\rm{m}}^{\rm{3}}}} \right) \times \left( {9.81\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}} \right) \times \left( {1250\;{\rm{m}}} \right)\\ \approx 15082.88\;{\rm{Pa}}\end{array}\)

The expression of the approximate force on the eardrum is given by,

\(F = \Delta PA\).

Here,\(F\)is the approximate force on the eardrum.

Substitute all the known values in the above expression.

\(\begin{array}{c}F = \left( {15082.88\;{\rm{Pa}}} \right) \times \left( {0.2 \times {{10}^{ - 4}}\;{{\rm{m}}^{\rm{2}}}} \right)\\ = 0.31\;{\rm{N}}\end{array}\)

Thus, the approximate force on the eardrum is \(0.31\;{\rm{N}}\).