Answer:
The density of water will be 0.935 g/[tex]cm^{3}[/tex]
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 [tex]\pi 0.6^{2}[/tex] = 1.130 [tex]cm^{2}[/tex]
The angle from the center of the dowel to the two water line edges is
T = (2)arccos)[tex]\frac{0.2}{0.6}[/tex]
So the area exposed above water is sector minus triangle:
[tex]\frac{0.60T}{2\pi -0.3}\sqrt{{0.6^{2} -0.4^{2}}[/tex]
= 0.642 - 0.447
= 0.195 [tex]cm^{2}[/tex]
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/[tex]cm^{3}[/tex]
