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1915]                    ON  THE  THEORY  OP THE  CAPILLARY  TUBE                        357
In the case of water a = 0'27 cm. If we take h0/a = O01, and cc/a = 4, we have dz/dx  0-OQS, so that (dz\dxf is still fairly small. Here for water 7i0 = 0'0027 cm. and 2# = 2*2 cm. A diameter of 2'2 cm. is thus quite insufficient, unless an error exceeding O'OOS cm. be admissible. Again, suppose h0/a = O'OOl, and take xja = 6. Then dz/dx = 0061, again small. For water 7i0=0-00027 cm., and 2=3'2 cm. This last value of h, is about that (O'OOSmm.) given by Richards and Coombs as the maximum admissible error of reading, and we may conclude that a diameter of 3'2 cm. is quite inadequate to take advantage of this degree of refinement.
We may go further in this example without too great a loss of accuracy. Retaining 7j0/a = 0-001, let us make #/a=7. I find 71(7) = 156 about, so that the extreme value of dzjdx is 0156, still moderately small. Here 2# = 3'8 cm., which is thus shown to be inadequate in the case of water.
But apart from the question of the necessary diameter of tube, information sufficient for experimental purposes can be derived in another manner. The initial value of z (on the axis) is 7*0; and z= 27i0 when / (#/a) = 2, i.e. when *'=r8. For the best work h0 should be on the limit of what caji be detected and then 7>0 and 2//0 could just be distinguished. The observer may be satisfied if no difference of level can be seen over the range x   l"8a; in the case of water this range is 2 x 1'8 x 0'27 = 097 cm., or say 1 cm.
It has already been remarked that when h0 is small enough x/a may become, great within the limits of application of (35). To shorten our expressions we will take a temporarily as the unit of length. Then when as is very great,
......................... (36)
Thus if ^r be the angle the tangent to the curve makes with the horizontal,
' ..................... (37)
an equation which may be employed when 7i0 is so small that a large oc is consistent with a small ty.
In order to follow the curve further, up to -^ = ^TT, we may employ the two-dimensional solution, the assumption being that the region of moderate ^ occupies a range of K small in comparison with its actual value, i.e. a value not much less than r, the radius of the tube. On account of the magnitude of x we have only the one curvature to deal with. For this curvature
-**-** Bin*-, ......................... (38)
jR     ds      dz
so that                             ^2 = G - cos i/r = 1 - cos -fy,..... (4)