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A capillary tube of radius 1 mm is held vertically with its lower end dipped in water (surface tension S = 0.075 N/m, contact angle theta = 0 deg). (a) Find the height to which water rises in the capillary. (b) If instead the tube length above the water is only half of the answer in part (a), find the new contact angle theta that the water surface makes with the tube wall.

1. (a) h = 1.5 cm, (b) theta = 60 deg
2. (a) h = 3.0 cm, (b) theta = 30 deg
3. (a) h = 1.5 cm, (b) theta = 30 deg
4. (a) h = 0.75 cm, (b) theta = 60 deg

Correct answer: (a) h = 1.5 cm, (b) theta = 60 deg

Solution

With theta = 0 the natural rise is h = 2S/(rho*g*r). If the tube is only h/2 long, water rises to the top but cannot overflow; instead the meniscus becomes less curved so the new contact angle theta' satisfies 2*S*cos(theta')/(rho*g*r) = h/2, i.e. cos(theta') = 1/2.

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