A weather forecasting plastic balloon of volume 15 m³ contains hydrogen of density 0.09 kg m^-3. The volume of an equipment carried by the balloon is negligible compared to it's own volume. The mass of empty balloon alone is 7.15 kg. The balloon is floating in air of density 1.3 kg m^-3. Calculate : (i) the mass of hydrogen in the balloon, (ii) the mass of hydrogen and balloon, (iii) the total mass of hydrogen, balloon and equipment if the mass of equipment is x kg, (iv) the mass of air displaced by balloon and (v) the mass of equipment using the law of floatation.

Given,

Volume of the plastic balloon = 15 m³

Density of hydrogen = 0.09 kg m^-3

Mass of the unfilled balloon = 7.15 kg

Density of air = 1.3 kg m^-3

(i) As we know,

Mass = volume of balloon x density of hydrogen

Substituting the value in the formula we get,

15 x 0.09 = 1.35 kg

Hence, mass of hydrogen in the balloon = 1.35 kg

(ii) Mass of hydrogen and balloon = mass of hydrogen in the balloon + mass of unfilled balloon = 7.15 + 1.35 = 8.5 kg

Hence, mass of hydrogen and balloon = 8.5 kg

(iii) Given,

mass of equipment = x kg

Hence,

Total mass of hydrogen, balloon and equipment = (8.5 + x) kg

(iv) Weight of the air displaced = upthrust = Volume of the balloon x density of the balloon x g

Hence,

Mass of the air displaced = volume of balloon x density of air = 15 x 1.3 = 19.5 kg

(v) By law of floatation,

Mass of the air displaced = total mass of hydrogen, balloon and equipment

Substituting the values in the formula we get,

$19.5 = 8.5 + x \\[0.5em] x = 19.5 - 8.5 \\[0.5em] x = 11 \text { kg} \\[0.5em]$

Hence, mass of the equipment = 11 kg