Chapter 10, Problem 70GP

To determine

To find: The approximate force on an eardrum of area.

Answer to Problem 70GP

The approximate force is $0.60\text{N}$ .

Explanation of Solution

Given:

The pressure at $0\text{m}$ is $101\text{kPa}$ .

At the $950\text{m}$ specified altitude, the pressure is $90\text{kPa}$ .

Formula used:

  $\text{Pressure}=\frac{F}{A}$

Here, $F$ is the force is applied,

  $A$ is the area of cross-section.

Calculation:

The pressure difference is given as,

  $\begin{array}{l}\Delta P=\rho gh=\frac{F}{A}\\ \Rightarrow \frac{F}{A}=\rho gh\Rightarrow F=A\rho gh\\ F={\left(0.50\times {10}^{-4}\text{m}\right)}^2\times \left(1.29\raisebox{1ex}{$\text{kg}$}\!\left/ \!\raisebox{-1ex}{${\text{m}}^{\text{3}}$}\right.\right)\times 9.80\raisebox{1ex}{$\text{m}$}\!\left/ \!\raisebox{-1ex}{${\text{s}}^{\text{2}}$}\right.\times 950\text{m}\\ F=0.60\text{N}\end{array}$

Conclusion:

The approximate force is $0.60\text{N}$ .