Question

Suppose that your body has a uniform density of$\mathbf{}\mathbf{0}\mathbf{.}\mathbf{95}$times that of water. (a) If you float in a swimming pool, what fraction of your body's volume is above the water surface? Quicksand is a fluid produced when water is forced up into sand, moving the sand grains away from one another so they are no longer locked together by friction. Pools of quicksand can form when water drains underground from hills into valleys where there are sand pockets. (b) If you float in a deep pool of quicksand that has a density $\mathbf{1}\mathbf{.}\mathbf{6}$times that of water, what fraction of your body's volume is above the quicksand surface? (c) Are you unable to breathe?

Short answer

Answer:

(a) Fraction of the body's volume above the water surface $=0.05$

(b) Fraction of the body's volume above the quicksand surface $=0.41$

(c) Yes, we can breathe though we are stuck in quicksand.

Step-by-step solution

Listing the given quantities

• The density of the body $\left({\rho }_b\right)=0.95$times the density of the water $\left({\rho }_w\right)$.
• The density of the quicksand $\left({\rho }_q\right)=6$times the density of the water $\left({\rho }_w\right)$.

Understanding the concept of buoyant force

By equating the buoyant force and weight of the fluid displaced by the body, we can find the relation ${\mathit{V}}_{\mathbf{f}}{\mathit{\rho }}_{\mathbf{f}}\mathbf{=}{\mathit{V}}_{\mathbf{b}}{\mathit{\rho }}_{\mathbf{b}}$. Using this, we can find the fraction of the body's volume above the fluid surface.

Formula:

Fraction $=\frac{Vb-Vf}{Vb}$

$=1-\frac{Vt}{Vb}$

$V_f{\rho }_f=V_b{\rho }_b$

(a) Calculation of fraction of the body's volume above the water surface

Fraction $=1-\frac{V_f}{V_b}$ (1)

Buoyant force $\left(F_b\right)=$weight of the fluid displaced by the body $\left(F_w\right)$

$V_f{\rho }_{f}g=V_b{\rho }_{b}g$

Rearranging the equation,

$\frac{Vf}{Vb}=\frac{\rho b}{\rho f}$

Using in above equation (1),

Fraction $=1-\frac{\rho b}{\rho f}$

$=1-\frac{0.95\rho W}{\rho w}$

$=1-0.05$

$=0.05$

Fraction of the body's volume above the water surface is$0.05$

(b) Calculation of fraction of the body's volume above the quicksand surface

Fraction $=1-\frac{\rho b}{\rho q}$

Using given values,

Fraction $=1-\frac{0.95\rho \mathrm{w}}{1.6\rho \mathrm{w}}$

$=1-0.593375$

$=0.4062$

$\approx 0.41$

Fraction of the body's volume above the quicksand surface is$0.41$

(c) Explanation for the calculation

From part $b,0.41$of the total human body will be above the quicksand. As the human body is less dense than quicksand, a person would sink to their chest and then float. Hence a person can breathe.