A weather forecasting plastic balloon of volume 15 m3 nbsp-contains hydrogen of density 0-09 kg m ndash-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 ndash-3-Calculate - nbsp-i- the mass of hydrogen in the balloon- nbsp-ii- the mass of hydrogen and balloon- nbsp-iii- the total mass of hydrogen- balloon and equipment if the mass of equipment is x kg- nbsp-iv- the mass of air displaced by balloon and nbsp-v- the mass of equipment using the law of floatation-

Dear #160;student(i) #160;weight #160;of #160;H2=density #160; #215; #160;volume #160;=0.09 #215;15=1.35(ii) #160;mass #160;of #160;H2+b ...

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Standard IX

Physics

Archimedes' Principle

A weather for...

Question

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.

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Solution

$D e a r s t u d e n t (i) w e i g h t o f H_{2} = d e n s i t y \times v o l u m e = 0.09 \times 15 = 1.35 (i i) m a s s o f H_{2} + b a l l o n = 1.35 + 7.15 = 8.5 k g (i i i) m a s s o f H_{2} + b a l l o o n + e q u i p p m e n t = 8.5 + x (i v) u p t h r u s t b y a i r = V ρ_{a i r} g = 15 \times 1.3 \times 10 = 195 N m a s s o f a i r d i s p l a c e d = 19.5 k g (v) u p t h r u s t b y a i r = V ρ_{a i r} g = 15 \times 1.3 \times 10 = 195 N w e i g h t o f H_{2} + b a l l o o n + e q u i p p m e n t = (8.5 + x) g = 195 x = 11 k g$

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