OF CAPILLARY ATTRACTION AND THE COHESION OF FLUIDS. 431 



which being multiplied by 3, becomes 



7- 2 -f3r=.0642. (316). 



544. Suppose now, that the observed altitude is 0.703 of an inch; 

 then, .by substituting 0.703 instead of h in equation (315), we obtain 



consequently, by completing the square, we get 



r+ 2.11r -4- 1.055 2 1.177225, 

 from which, by extracting the square root, we obtain 



r + 1. 055 = 1.085; 



hence, by subtraction, we have 



r 1.085 1.055 = . 03 of an inch. 



Now, if one third of the radius just found, be added to the observed 

 altitude, the sum thence arising will express the mean altitude ; and 

 if the whole radius be added to the observed altitude, the sum will 

 express the greatest height to which the fluid rises in the tube. 



PROBLEM LXX. 



545. Two parallel planes of glass or other materials, are placed 

 in a vertical position, with the lower sides immersed in a fluid : 



It is required to determine how high the fluid rises between 

 them, their distance asunder being very small in comparison 

 to their surfaces. 



Let ad and be represent the ends or sections of two plates of glass, 

 placed in a position of vertical parallelism, 

 and having their lower edges d and c im- 

 mersed in a fluid of which the surface is IK. 

 Suppose now that a b or cd, the distance 

 between the plates, is very small in com- 

 parison to their extent of surface ; then it is 

 obvious, that the fluid will rise between them 

 as high as e in consequence of the immersion, 

 and from thence it moves upwards by the 

 attractive influence of the glass, and the 



mutual action of its own particles, until it arrives at rs, where it forms 

 a semi-cylindrical meniscus rsfg, whose diameter is equal to the dis- 

 tance between the planes, and its length the same as their horizontal 

 breadth. 



