Chapter 10, Problem 75GP

To determine

To calculate: The height of the water that can be sucked up the straw.

Answer to Problem 75GP

The height of the water is found to be $1.08\text{m}$ .

Explanation of Solution

Given:

The decreased amount of pressures is, $P=-80\text{mm-Hg}$

  $d_w=$ the density of the water $=1\times {10}^{-3}{\text{g·cm}}^{\text{-3}}$

Formula used:

The pressure difference is, $\Delta P=\rho g\Delta h$

Where, $\rho$ is the density,

g is the acceleration gravity,

and $\Delta h$ change in water level.

Calculation:

The pressure change in column of water is the is the pressure difference due to the lungs. So, the height is given as,

  $\begin{array}{l}\Delta h=\frac{\Delta P}{\rho g}=\frac{80\text{mm-Hg}\left(\frac{133\raisebox{1ex}{$\text{N}$}\!\left/ \!\raisebox{-1ex}{${\text{m}}^{\text{2}}$}\right.}{1\text{mm-Hg}}\right)}{1.00\times {10}^3\raisebox{1ex}{$\text{kg}$}\!\left/ \!\raisebox{-1ex}{${\text{m}}^{\text{3}}$}\right.\times 9.80\raisebox{1ex}{$\text{m}$}\!\left/ \!\raisebox{-1ex}{${\text{sec}}^{\text{2}}$}\right.}\\ =1.08\text{m}\end{array}$

Conclusion:

The height of the water is found to be $1.08\text{m}$ .