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A hollow metallic sphere floats in water so that its topmost point just reaches the water surface (fully submerged). If d is the density of the metal (in cgs units, i.e. relative to water = 1 g/cc), what fraction of the sphere's total volume is the hollow (empty) part?
- 1/d
- (1 - 1/d)
- d/(d - 1)
- (1 + 1/d)
Correct answer: (1 - 1/d)
Solution
Let total volume be V and hollow volume be v. Volume of metal = V - v. Since the sphere just floats fully submerged, weight of metal = weight of water displaced by full volume V. In cgs with water density 1: d*(V - v) = 1*V. So d*V - d*v = V, giving d*v = (d - 1)V, hence v/V = (d - 1)/d = 1 - 1/d.
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