MECHANICAL STRETCHING OF LIQUIDS. 
367 
subjected to pressures of 10 and 12 atmospheres (above the external atmospheric 
pressure), and the retreat of the end of the mercury column along the tube was 
noted. The following measures of this recession were obtained for a pressure of 10 
gauge-atmospheres of 15 pounds per sq. inch. 
millims 
Direct observation at 10 atmospheres . 
. . 7-31 
7-33 
From observation at 12 atmospheres 
. . 7'40 
7'40 
7-1 
Mean .... 
. . 7'31 
The mean value is equivalent to 7'22 millims. per 10 atmospheres of 1033 grms. 
per sq. centim. 
The measures were somewhat unsatisfactory on account of (1) the smallness of the 
total length to be measured, (2) the deformation of the meniscus due partly to a 
certain want of cleanness at the surface which dragged along the glass, (3) a 
tendency to entangle bubbles of alcohol between the mercury and the sides of the 
tube owing to the churning as the end of the column moved to and fro. This ulti¬ 
mately caused a separation of the column into segments, and must have had the 
effect of diminishing the apparent recession. (4) It should also be mentioned that 
the end of the column was situate in the upper part of the tube FE, which, owing to 
the subsequent breakage of the apparatus, I was not able to specially calibrate, but 
which, if we may judge from the measures of the bubbles, was probably somewhat 
wider than the part below the wire. On this account also the measured result is 
probably rather too small. 
If the mercury were quite incompressible, and if there were no residual alcohol in 
the vessel, this number would represent, in millims. of the tube, the yielding of the 
glass per 10 atmospheres, but by reason of the corrections required on these two 
accounts this length is reduced to 6’626 millims.' 5 ' 
* The calculation of these corrections was made as follows :—Let the change required in the volume 
of the glass per 10 atmospheres, be v g millims. of the tube, and let the corresponding changes in 
the volume of the mercury, and of the residual alcohol be v m and v a millims. of the tube respectively, and 
let the observed apparent alteration of volume be (a) millims. of tube. Then 
v ff + v m + v a = a .(i.). 
Let the vessel, when full, contain n times as much alcohol as this residue. Then admitting extension and 
compression under numerically equal stresses to have been proved equal (p. 367), we have 
(»•), 
v,j + nv a = A 
