Question

An empty rubber balloon has a mass of 12.5 g. The balloon is filled with helium at a density of 0.181 kg/m3. At this density the balloon has a radius of 0.498 m. If the filled balloon is fastened to a vertical line, what is the tension in the line? The density of air is 1.29 kg/m3.

Answer 1

Answer:

• 5.5 N

Explanation:

mass of balloon (m) = 12.5 g = 0.0125 kg

density of helium = 0.181 kg/m^{3}

radius of the baloon (r) = 0.498 m

density of air = 1.29 kg/m^{3}

acceleration due to gravity (g) = 1.29 m/s^{2}

find the tension in the line

the tension in the line is the sum of all forces acting on the line

Tension =buoyant force  + force by helium + force of weight of rubber

force = mass x acceleration

from density = \frac{mass}{volume} ,  mass = density x volume

• buoyant force =  density x volume x acceleration

        where density is the density of air for the buoyant force

        buoyant force = 1.29 x (\frac{4]{3} x π x 0.498^{3}) x 9.8 = 6.54 N

• force by helium =  density x volume x acceleration

        force by helium =  0.181 x (\frac{4]{3} x π x 0.498^{3}) x 9.8 = 0.917 N

• force of its weight = mass of rubber x acceleration

        force of its weight = 0.0125 x 9.8 = 0.1225 N

• Tension = buoyant force  + force by helium + force of weight of rubber

         the force  of weight of rubber and of helium act downwards, so they      

          carry a negative sign.

• Tension = 6.54 - 0.917 - 0.1225 = 5.5 N

Answer 2

An empty rubber balloon has a mass of 12.5 g. The balloon is filled with helium at a density of 0.181 kg/m3. The balloon is filled with helium at a density of 0.181 kg/m3. The tension in the line is 5.5 N

The tension in the line is relatively equal to the difference between the  buoyant force in the upward direction and the weight of the balloon together with the helium acting in the downward direction

It is given by the formula:

$\mathbf{T = F_b -W_{helium}- W_{ballon}}$

where;

• $\mathbf{F_b = buoyant \ force}$
• $\mathbf{ W_{helium} = weight \ of \ the \ helium }$
• $\mathbf{ W_{ballon} = weight \ of \ the \ ballon }$

To start with the buoyant force that can be computed by using the formula:

$\mathbf{F_b = \rho_{air} \times v_g}$

where;

• v = volume of the ballon = $\mathbf{\dfrac{4}{3}\pi r^3}$

$\mathbf{F_b = \rho_{air} \times \dfrac{4}{3} \pi r^3 g}$

$\mathbf{F_b = 1.29 \times \dfrac{4}{3} \pi (0.498)^3 \times 9.8}$

$\mathbf{F_b = 6.540 \ N}$

Similarly, the weight of the ballon:

$\mathbf{W_{helium }= m_{helium } \times g}$

Recall that;

$\mathbf{density = \dfrac{mass}{volume}}$

mass = density $\mathbf{\rho}$ × volume (v)

∴

$\mathbf{W_{helium }=\rho_{helium } \times v \times g}$

$\mathbf{W_{helium }=0.181 \times \dfrac{4}{3}\pi (0.498)^3 \times 9.8}$

$\mathbf{W_{helium }=0.918 \ N}$

Also, the weight of the ballon is calculated as:

$\mathbf{W_{ballon }=m_{ballon}\times g}$

$\mathbf{W_{ballon }=0.0125 \ kg \times 9.8 \ m/s^2}$

$\mathbf{W_{ballon }=0.1225 \ N}$

Therefore, the tension in the line can be computed as:

$\mathbf{T = F_b -W_{helium}-W_{ballon}}$

$\mathbf{T = 6.540 \ N -0.918 \ N-0.1225 \ N}$

T = 5.4995 N

T ≅ 5.5 N

Therefore, we can conclude that the  tension in the line is 5.5 N

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