Question

A balloon of volume 120 m³ is filled with hot air, of density 38 kg^-3. If the fabric of balloon weighs 12 kg, such that additional equipment of wt. x is attached to it, calculate the magnitude of Density of cold air is 1.30 kgm^-3.

Answer 1

Volume of balloon = V = 120 m³

Density of hot air = ${\displaystyle {\rho }_{\text{hot air}}}$ = 0.38 kgm^-3

Mass of empty balloon = 12 kg

Weight of the empty balloon = 12 kgf

Weight of the additional equipment attached with the balloon

=x kgf

Density of cold air = ${\displaystyle {\rho }_{\text{cold air}}}$ = 1.30 Kgm³

Volume of balloon=Volume of hot air inside the balloon=Volume of cold air displaced by balloon

= V = 120 m³ Weight of hot air = ${\displaystyle \text{V}{\text{p}}_{\text{hot air}}}$ g

= 120 x 0.38 x g = 45.6kgf

Weight of empty balloon + Weight of hot air inside the balloon + Weight of equipment = Downthrust

12 + 45.6 + x = Downthrust

Downthrust = 57.6 + x

Upthrust = Weight of cold air diplaced by balloon

= ${\displaystyle \text{V}\text{ρ}\text{cold air}}$ g

= 120 x 1.30 xg= 156kgf

By law of floatation:

Downthrust = Upthrust

57.6 + x = 156

x = 156-57.6

x = 98.4 kgf