Question

(II) How many helium-filled balloons would it take to lift a person? Assume the person has a mass of $72\mathrm{k}\mathrm{g}$and that each helium-filled balloon is spherical with a diameter of$33\mathrm{c}\mathrm{m}$.

Short answer

The number of balloons required is $3472.6\mathrm{b}\mathrm{a}\mathrm{l}\mathrm{l}\mathrm{o}\mathrm{o}\mathrm{n}\mathrm{s}$ to lift a person.

Step-by-step solution

Concept of volume density

Volume density is the volume of an object per unit density of that object. It is used to deduce the number of particles in a space.

Given data

The mass of the person is $\mathrm{m}=72\mathrm{k}\mathrm{g}$.

The diameter of the helium-filled balloon is $d=33\mathrm{c}\mathrm{m}$.

Calculation of the buoyant force and density

The buoyant force is calculated as:

$F_{Bouyant}=\rho Vg$

Here,$\rho$is density of fluid, $V$ is volume occupied and $\mathrm{g}$ is acceleration due to gravity.

The density of a fluid is calculated as:

$\rho =\frac{m}{V}$

Here,$m$is the mass of the fluid.

Using law of equilibrium of forces, buoyant force will be equal to the force due to weight $F_{\mathrm{w}\mathrm{e}\mathrm{i}\mathrm{g}\mathrm{h}\mathrm{t}}$.

$\begin{array}{rl}{F}_{\mathrm{b}\mathrm{o}\mathrm{u}\mathrm{y}\mathrm{a}\mathrm{n}\mathrm{t}}& =F_{\mathrm{w}\mathrm{e}\mathrm{i}\mathrm{g}\mathrm{h}\mathrm{t}}\\ {\rho }_{\mathrm{a}\mathrm{i}\mathrm{r}}{V}_{\mathrm{H}\mathrm{e}}g& =mg+{\rho }_{\mathrm{H}\mathrm{e}}{V}_{\mathrm{H}\mathrm{e}}g\\ V_{\mathrm{H}\mathrm{e}}& =\frac{m}{{\rho }_{\mathrm{a}\mathrm{i}\mathrm{r}}-{\rho }_{\mathrm{H}\mathrm{e}}}\end{array}$

Here, ${\rho }_{\mathrm{a}\mathrm{i}\mathrm{r}}$is the density of air, ${\rho }_{\mathrm{H}\mathrm{e}}$ is the density of Helium and$V_{\mathrm{H}\mathrm{e}}$ is the volume of helium.

Calculation of capacity to lift

The volume of a Helium-filled balloon is:

$V_{\mathrm{H}\mathrm{e}}=N\frac{4}{3}\pi r^3$

Here, N is the number of balloon and r is the radius.

Substituting the values in the above equation.

$\begin{array}{rl}N\frac{4}{3}\pi r^3& =\frac{m}{{\rho }_{\mathrm{a}\mathrm{i}\mathrm{r}}-{\rho }_{\mathrm{H}\mathrm{e}}}\\ N& =\frac{3m}{4\pi r^3\left({\rho }_{\mathrm{a}\mathrm{i}\mathrm{r}}-{\rho }_{\mathrm{H}\mathrm{e}}\right)}\end{array}$

Substitute the known values in the above equation to find the number of helium-filled balloon.

$\begin{array}{rl}N& =\frac{3\times 72\mathrm{k}\mathrm{g}}{4\pi {\left(\frac{33\mathrm{c}\mathrm{m}\times \frac{1\mathrm{m}}{100\mathrm{c}\mathrm{m}}}{2}\right)}^3\left(1.29\mathrm{k}\mathrm{g}/{\mathrm{m}}^3-0.179\mathrm{k}\mathrm{g}/{\mathrm{m}}^3\right)}\\ N& =3472.6\mathrm{b}\mathrm{a}\mathrm{l}\mathrm{l}\mathrm{o}\mathrm{o}\mathrm{n}\mathrm{s}\end{array}$

Hence, the number of balloons are $3472.6\mathrm{b}\mathrm{a}\mathrm{l}\mathrm{l}\mathrm{o}\mathrm{o}\mathrm{n}\mathrm{s}$.