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A small sphere is released from rest and falls through air. After descending 100 m it reaches terminal velocity and thereafter falls at constant speed. Considering the first 100 m and the next 100 m, which statement about the magnitudes of work done is correct?
- by viscosity of air is lesser in first 100 m than in the second 100 m
- by buoyancy of air is in first 100 m is equal to that in the second 100 m
- by viscosity of air is greater in first 100 m than in the second 100 m
- by buoyancy of air is lesser in first 100 m than that in the second 100 m
Correct answer: by buoyancy of air is in first 100 m is equal to that in the second 100 m
Solution
Buoyancy depends only on the sphere's volume and air density, so it is the same constant force throughout. Over equal 100 m distances it does equal work. Viscous force grows with speed, so it is smaller during the accelerating first 100 m than during the constant-speed second 100 m.
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