A weather forecasting plastic balloon of volume $15 m^{3}$ contains hydrogen of density $0.09 k g$ $m^{- 3}$ . The volume of an equipment carried by the balloon is negligible compared to its own volume. The mass of empty balloon along is $7.15 k g$ . The balloon floating in air of density $1.3 k g$ $m^{- 3}$ . 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 dispalced by balloon, and (v) the mass of equipment using the law of floatation.

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Solution

Mass of hydrogen $= P \times V = .15 \times 0.09$
$= 1.35 k g$
Total mass $=$ mass of $H + B a l l o o n$
$= 1.35 + 7.15$
$= 8.50 k g$
Mass of hydrogen $= B a l l o o n + f l o a t i n g$
$e q u i p m e n t = 1.35 + 7.15 + 8.50 m$
Mass of displaced $= 15 \times 1.3 = 19.5 k g$
So $8.5 + x = 19.5$
$x = 11 k g$ .

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Mass of hydrogen =P×V=.15×0.09=1.35 kgTotal mass = mass of H+Balloon=1.35+7.15=8.50 kgMass of hydrogen =Balloon+floatingequipment=1.35+7.15+8.50 mMass of displaced =15×1.3=19.5 kgSo 8.5+x=19.5x=11 kg.