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A toy balloon, which has a mass of 2.90 g before it is inflated, is filled with helium (with a density of 0.180 kg/m^3) to a volume of 8400 cm^3. What is the minimum mass that should be hung from the balloon to prevent it from rising up into the air? Assume the air has a density of 1.29 kg/m^3.

User Anuj Gupta
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1 Answer

5 votes

Answer:


M=6.4243\ g

Step-by-step explanation:

Given:

  • mass of deflated balloon,
    m_b=2.9\ g=0.0029\ kg
  • density of helium,
    \rho_h=0.180\ kg.m^(-3)
  • volume of inflation,
    V=8400\ cm^3=0.0084\ m^3
  • density of air,
    \rho_a=1.29\ kg.m^(-3)

To stop this balloon from rising up we need to counter the buoyant force.

mass of balloon after inflation:


m=m_h+m_b


m=0.0084* 0.180+0.0029


m=0.004412\ kg

Now the density of inflated balloon:


\rho_b=(m)/(V)


\rho_b=(0.004412)/(0.0084)


\rho_b=0.5252\ kg.m^(-3)

Now the buoyant force on balloon


F_B=V(\rho_a-\rho_b).g


F_B=0.0084(1.29-0.5252)* 9.8


F_B=0.063\ N

∴Mass to be hung:


M=(F_B)/(g)


M=0.00642432\ kg


M=6.4243\ g

User Armali
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