(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

Expert verified

The number of balloons required is \(3472.6\;{\rm{balloons}}\) to lift a person.

Step by step solution

01

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.

02

Given data

The mass of the person is \({\rm{m}} = 72\;{\rm{kg}}\).

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

03

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.

04

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 \(3472.6\;{\rm{balloons}}\).

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Figure 10-53

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