## Related questions with answers

(a) A 75.0-kg man floats in freshwater with 3.00% of his volume above water when his lungs are empty, and 5.00% of his volume above water when his lungs are full. Calculate the volume of air he inhales-called his lung capacity-in liters. (b) Does this lung volume seem reasonable?

Solution

Verified$\textbf{(a)}$ Here the mass of the man is

$\begin{align*} m = 75.0\,\mathrm{kg} \end{align*}$

$\textbf{Case-I:}$ The fraction of the volume of the man above water when his lungs are empty is

$\begin{align*} f' = 3\,\% = 0.03 \end{align*}$

Therefore the fraction submerged is

$\begin{align*} f = 97\,\% = 0.97 \end{align*}$

Hence from the equation 11.41, the defination of fraction submerged is

$\begin{align*} f & = \frac{\rho_{\text{man}}}{\rho_{\text{warer}}} \\ \end{align*}$

Hence, we get

$\begin{align*} \rho_{\text{man}} & = f\, \rho_{\text{water}}\\ & = 0.97 \times\,1.000\times 10^3\,\mathrm{kg\,m^{-3}}\\ & = 0.97\times 10^3\,\mathrm{kg\,m^{-3}} \end{align*}$

Hence, the volume of the body of the man is

$\begin{align*} V_1 & = \frac{m}{\rho_{\text{man}}} \\ & = \frac{75.0\,\mathrm{kg}}{0.97\times 10^3\,\mathrm{kg\,m^{-3}}}\\ & = 77.32\times 10^{-3}\,\mathrm{m^{3}}\tag{1} \end{align*}$

$\textbf{Case-II:}$ The fraction of the volume of the man above water when his lungs are full with air is

$\begin{align*} f' = 5\,\% = 0.05 \end{align*}$

Therefore the fraction submerged is

$\begin{align*} f = 95\,\% = 0.95 \end{align*}$

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