Question

please show all your work! thanks

A capillary tube with a radius of mm is inserted vertically into a container of water. As shown in the picture at right, water wets the inner wall of the capillary with a contact angle of 0°. The surface tension force, Fst, (or capillary force in this context) draws water up the tube until the meniscus reaches a level of hw above the water level in the container. You may assume that surface tension of water, y-72.9 mN/m, density of water, p 1000 kg/m2, and g 10 m/s. (a) [2 points] Find Fxt (b) 12 points] Find w. The capillary tube is taken out of water and carefully dried to be rid of water. Then it is inserted vertically into a container of mercury. As shown in the picture at right, the surface tension force, Fst, pushes the mercury meniscus down below the mercury level in the container. You are given that the contact angle between mercury and the capillary wall is 140°, surface tension of mercury, ym-486.5 mN/m, and density of mercury, pm 13,600 kg/m. 140 (c) [3 points] Find Fst (d) [3 points] Find h

Answer 1

The liquid (water) continue to rise in the capillary tube until the weight of the liquid column becomes equal to force due to surface due to surface tension Surface tension force F_st = gamma(2 times pi r) times cos theta weight of liquid (water rise) mg = (rho pi r^2 h)g F_st = mg gamma(2 times pi r) times cos theta = (rho pi r^2 h)g h = 2 gamma cos theta/rho g r a) F_st = gamma_w (2 times pi r) times cos theta_w F_st = 72.9 times 10^-3 times (2 times pi times 10^-3) times cos 0 = 4.58 times 10^-4 N F_st = 0.458 mN b) h_w = 2 gamma_w cos theta_w/rho_w gr h_w = 2 times 72.9 times 10^-3 times cos 0/1000 times 9.8 times 10^-3 = 14.8 times 10^-3 m h_w = 14.8 mm c) \r\n when capillary tube is dipped into mercury F_st = gamma_m (2 times pi r) times cos theta_m