Question

1009 cm
9. 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 its 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.​

Answer 1

Given:-

• Volume of plastic balloon = 15m³.

• Mass of empty balloon = 7.15 kg.

• Density of hydrogen = 0.09 kg/m³

• Density of air = 1.3 kg/m³

Answer:-

(1) The mass of  hydrogen in the balloon = V × ρ  

Where,V = volume & ρ is density.

➫ 15 × 0.09kg = 1.35 kg.

∴ The mass of  hydrogen in the balloon is 1.35kg.

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(2) The mass of hydrogen and  balloon=Mass of balloon + mass of H

➫ 1.35 + 1.75 = 8.50 kg.

∴ The mass of hydrogen and  balloon = 8.5 kg.

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(3) The total mass of hydrogen, balloon and  equipment if the mass of equipment is x kg is (8.5 + x )kg.

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(4) Mass displaced = V × ρ  

⇒  15×1.3=19.5kg.

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(5) Using law of floation to find mass of equipment :-

Mass of air displaced = Mass of H + ballon + equipment

⇒ 19.5 = 8.5 + x

⇒ x = 11 kg.

∴ Mass of equipment is 11kg.

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