Question

An empty rubber balloon has a mass of 0.0105 kg. The balloon is filled with helium at a density of 0.181 kg/m3. At this density the balloon is spherical with a radius of 0.300 m. If the filled balloon is fastened to a vertical line, what is the tension in the line?

Answer 1

If the buoyancy force is neglected tension in the line is due to weight of balloon and helium

mass of the rubber balloon (M_b = 0.0105kg

density of helium () = 0.181 kg/m³

radius of the balloon (r) = 0.300 m

Tension in the line = weight of balloon+ weight of helium

   weight of the balloon = M_bg =0.0105 (9.8) = 0.1029 N

   weight of the helium gas = (mass of helium)(g)

   = (volume *density) g

   =(((4/3)r³ * ) g

   =((4/3) (0.3)³ )(0.181)(9.8)

   = 0.20061 N

Tension in the wire = weight of balloon + weight of the helium

= 0.1029+0.2006

= 0.3035 N

If buoyancy is considered, tension in wire = weight of balloon + weight of helium - buoyancy force

buoyancy force=(density of air )(Volume of balloon )(g)

density of air =1.225 kg/m^3

   Buoyancy force = 1.225((4/3)0.3³)9.8

= 1.3577

Tension in the line = 0.3035 - 1.3577 =- 1.0542 N

   1.0542 will be acted upwards