hi guys, need some help for this question.(adsbygoogle = window.adsbygoogle || []).push({});

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.

**Physics Forums - The Fusion of Science and Community**

# Capillary action: mercury depression

Know someone interested in this topic? Share a link to this question via email,
Google+,
Twitter, or
Facebook

- Similar discussions for: Capillary action: mercury depression

Loading...

**Physics Forums - The Fusion of Science and Community**