Physics Help Forum capillary action: mercury depression
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 May 1st 2009, 02:19 PM #1 Junior Member   Join Date: May 2009 Posts: 2 capillary action: mercury depression hi guys, need some help for this question. a capillary of unknown internal radius was inserted into a pool of mercury into a pool of mercury. the height of mercury within the capillary was depressed 1.6cm below the free liquid. Calculate the internal diameter of the capillary. For the mercury-air interface on glass, the contact angle is 140 degree. for mercury at 298.15K, density= 13.59g/cm3, surface tension=0.4865Nm-1 im not sure how to go about solving this question. is it the same approach for capillary rise, but in this case it's a depression, which is due to the cohesive force being greater than the adhesive force. for capillary rise, 2 x pi x r x surface tension x cos angle = pi x r2 x h x density x g.
 May 9th 2009, 10:31 PM #2 Senior Member   Join Date: Apr 2008 Location: HK Posts: 886 I guess the same approach is used. Simply the difference is that the cosine of the contact angle is negative and the distance h should be the distance between the top of meniscus and free liquid water level. Last edited by werehk; May 9th 2009 at 10:36 PM.

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