A capillary tube of radius r is immersed in water and water rises in it to a height h. The mass of the water in the capillary is 5 g. Another capillary tube of radius 2r is immersed in water. The mass of water that will rise in this tube is:

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(2) 10.0 g. Given: m₁=5 g in a tube of radius r; second tube of radius 2r, same water. Idea: the weight of the raised column is held up by the surface tension pulling along the circle of contact, so the mass follows the circumference, not the area. 2π rT cos θ=mg, so m=(2π rT cos θ)/g. With T, θ and g fixed, m∝ r. Check by the other route: h=(2T cos θ)/(rρ g) and m=ρπ r²h, so the r² and the 1/r leave one power of r. (m₂)/(m₁)=(2r)/r=2 m₂=2×5.0=10.0 g

A capillary tube of radius r is immersed in water and water rises in it to a height h. The mass of the water in the capillary is 5 g. Another capillary tube of radius 2r is immersed in water. The mass of water that will rise in this tube is:

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