Question

When you ascend or descend a great deal when driving in a car, your ears “pop,” which means that the pressure behind the eardrum is being equalized to that outside? If this did not happen, what would be the approximate force on an eardrum of area \(0.20\;c{m^2}\)if a change in altitude of 1250 m takes place?

Short answer

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

Step-by-step solution

Understanding the pressure

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.

Identification of given data

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

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

Determining the pressure difference

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.

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

Determining the approximate force on the eardrum

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 \(0.31\;{\rm{N}}\).