Question

(II) How many helium-filled balloons would it take to lift a person? Assume the person has a mass of \({\rm{72 kg}}\)and that each helium-filled balloon is spherical with a diameter of\(33\;{\rm{cm}}\).

Short answer

The number of balloons required is \(3472.6\;{\rm{balloons}}\) 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 \({\rm{m}} = 72\;{\rm{kg}}\).

The diameter of the helium-filled balloon is \(d = 33\;{\rm{cm}}\).

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 \({\rm{g}}\) 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 \({F_{{\rm{weight}}}}\).

\(\begin{aligned}{F_{{\rm{bouyant}}}} &= {F_{{\rm{weight}}}}\\{\rho _{{\rm{air}}}}{V_{{\rm{He}}}}g &= mg + {\rho _{{\rm{He}}}}{V_{{\rm{He}}}}g\\{V_{{\rm{He}}}} &= \frac{m}{{{\rho _{{\rm{air}}}} - {\rho _{{\rm{He}}}}}}\end{aligned}\)

Here, \({\rho _{{\rm{air}}}}\)is the density of air, \({\rho _{{\rm{He}}}}\) is the density of Helium and\({V_{{\rm{He}}}}\) is the volume of helium.

Calculation of capacity to lift

The volume of a Helium-filled balloon is:

\({V_{{\rm{He}}}} = N\frac{4}{3}{\rm{\pi }}{r^3}\)

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

Substituting the values in the above equation.

\(\begin{aligned}N\frac{4}{3}{\rm{\pi }}{r^3} &= \frac{m}{{{\rho _{{\rm{air}}}} - {\rho _{{\rm{He}}}}}}\\N &= \frac{{3m}}{{4{\rm{\pi }}{r^3}\left( {{\rho _{{\rm{air}}}} - {\rho _{{\rm{He}}}}} \right)}}\end{aligned}\)

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

\(\begin{aligned}N &= \frac{{3 \times 72\;{\rm{kg}}}}{{4\pi {{\left( {\frac{{33\;{\rm{cm}} \times \frac{{1\;{\rm{m}}}}{{100\;{\rm{cm}}}}}}{2}} \right)}^3}\left( {1.29\;{\rm{kg/}}{{\rm{m}}^3} - 0.179\;{\rm{kg/}}{{\rm{m}}^3}} \right)}}\\N &= 3472.6\;{\rm{balloons}}\end{aligned}\)

Hence, the number of balloons are \(3472.6\;{\rm{balloons}}\).