A balloon of volume 120 m^{3} 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}.

#### Solution

Volume of balloon = V = 120 m^{3}

Density of hot air = `ρ_"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 = `ρ_"cold air"` = 1.30 Kgm^{3}

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

= V = 120 m^{3 }Weight of hot air = `"V""p"_"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

= `"V""ρ""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