Answer:
The density of water will be 0.935 g/
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Step-by-step explanation:
It is sufficient to solve this as a 2D problem since the length of the dowel does not matter. Let p be the density of the dowel in g/cm^3, and take water density 1g/cm^{3}
The area of the whole circle is
= 1.130
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The angle from the center of the dowel to the two water line edges is
T = (2)arccos)
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So the area exposed above water is sector minus triangle:
= 0.642 - 0.447
= 0.195
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So, the weight of the dowel is balanced by the weight of water displaced:
1.130 p
= (1.130 - 0.195)
p = 0.935 g/
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