Question

A helium-filled balloon, whose envelope has a mass of 0.28 kg, is tied to a 3.5-m long, 0.051-kg string. The balloon is spherical with a radius of 0.40 m. When released, it lifts a length h of the string and then remains in equilibrium, as in the figure below. Determine the value of h. Hint: Only that part of the string above the floor contributes to the load being supported by the balloon. (The density of air is 1.29 kg/m3 and the density of helium is 0.179 kg/m3.) He

Answer 1

Radius of the spherical balloon, r = 0.40 m

So, volume of the spherical balloon,

V = 4/3 π r^3

= 4/3 π (0.40 m)^3 = 0.268 m^3

Now, density of helium = 0.179 kg/m^3.

So, weight of helium in the balloon = 0.179 x 0.268 = 0.048 kg.

Given that weight of the balloon = 0.28 kg

Total weight of balloon with helium = 0.28 kg + 0.048 kg = 0.328 kg

Density of air = 1.29 kg/m^3

So, weight of air displaced = 0.268 m^3 × 1.29 kg/m^3 = 0.346 kg

Therefore,

Lifting mass on the balloon = 0.346 kg - 0.328 kg

= 0.018 kg

Length per unit mass of the string = 3.5 / 0.051 m/kg

Therefore, length of lifted string at the balanced condition of the balloon = 0.018 kg x (3.5 / 0.051 kg/m)

= 1.23 m (Answer)