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A solid sphere of mass m and radius r lies at the bottom of a deep tank of water of depth h. The sphere's material has exactly the same density as water. The sphere is then very slowly raised completely out of the water. How much work is done by the external agent in lifting the sphere out?
- mgh
- 0.5 mgh
- mg(0.5r + h)
- mg(r + h)
Correct answer: mg(0.5r + h)
Solution
Because the densities match, weight and buoyant force cancel while the sphere is wholly under water, so no net work is needed over the depth h apart from raising it; the agent must do positive work only during emergence, where the average extra force over the rise of one diameter equals half the weight.
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