The surface tension of soapy water is $0.0250 \mathrm{~N} / \mathrm{m}$. What is the height of the water in a capillary tube of diameter $2.00 \mathrm{~mm}$ ?
A) $5.10 \mathrm{~mm}$
B) $4.21 \mathrm{~mm}$
C) $3.98 \mathrm{~mm}$
D) $4.77 \mathrm{~mm}$
E) $3.66 \mathrm{~mm}$

Question

Quizplus · Accepted Answer

The height of the water in the capillary tube is given by:$$h = \frac{2 \gamma \cos \theta}{\rho g r}$$where:* $\gamma$ is the surface tension of the water* $\theta$ is the contact angle between the water and the glass* $\rho$ is the density of the water* $g$ is the acceleration due to gravity* $r$ is the radius of the capillary tubeIn this problem, we are given that:* $\gamma = 0.0250 \mathrm{~N} / \mathrm{m}$* $\theta = 0^\circ$ (since the water is wetting the glass)* $\rho = 1000 \mathrm{~kg} / \mathrm{m}^3$* $g = 9.81 \mathrm{~m} / \mathrm{s}^2$* $r = 1.00 \mathrm{~mm} = 0.001 \mathrm{~m}$Plugging these values into the equation, we get:$$h = \frac{2 (0.0250 \mathrm{~N} / \mathrm{m})(1)}{\rho g r} = \frac{2 (0.0250 \mathrm{~N} / \mathrm{m})(1)}{1000 \mathrm{~kg} / \mathrm{m}^3 (9.81 \mathrm{~m} / \mathrm{s}^2) (0.001 \mathrm{~m})} = 5.10 \mathrm{~mm}$$Therefore, the height of the water in the capillary tube is 5.10 mm.

The Surface Tension of Soapy Water Is 0.0250 N/m0.0250 \mathrm{~N} / \mathrm{m}0.0250 N/m

The Surface Tension of Soapy Water Is $0.0250 \mathrm{~N} / \mathrm{m}$