Question

Mercury has a surface tension coefficient σ = 0.487N/m. A Mercury barometer has a cylindrical tube that is 5mm in diameter.

a) By how many mm will the Mercury level change due to capillary forces? Follow the procedures and assumptions of the example in the notes.

b) Air pressure at sea level typically causes Mercury level to rise by 760mm. What is the percent error due to capillary force?

c) Repeat parts a and b for a tube that is 0.1mm in diameter.

d) Repeat parts a and b for a tube that is 20mm in diameter.

Answer 1

a) here the formulae to calculate the raise of mercury in cylindrical tube (h)

assume angle of contact is 0 degree

h = 2Tcos0/ rdg

here T = 0.487 N/m

angle of contact = 0 degree

r is the radius of tube = Diameter/2 = 5 * 10^-3 / 2 =  2.5 * 10^-3 m

d id density of mercury = 13546 kg / m³

g is accelerqation due to gravity 9.8 m/s²

therefore h = 2*0.487*1/ 2.5 *10^-3*13546*9.8 = +2.934 mm

b) therefore here as we have a positive error of  +2.934 mm due to capillary force

so air pressure is 760 +2.934 mm = 762.934 mm

error = 760 -  762.934 /760 = 0.38% error.

c)

h = 2Tcos0/ rdg

here T = 0.487 N/m

angle of contact = 0 degree

r is the radius of tube = Diameter/2 = 0.1 * 10^-3 / 2 = 5* 10^-5 m

d id density of mercury = 13546 kg / m³

g is accelerqation due to gravity 9.8 m/s²

therefore h = 2*0.487*1/ 5* 10^-5*13546*9.8 = 0.146 m = +146.74 mm

therefore here as we have a positive error of  +146.74 mm due to capillary force

so air pressure is 760 + 146.74 mm = 906.74 mm

error = 760 - 906.74 /760 = 19.3%error.

d)

h = 2Tcos0/ rdg

here T = 0.487 N/m

angle of contact = 0 degree

r is the radius of tube = Diameter/2 = 20 * 10^-3 / 2 = 10 * 10^-3m

d id density of mercury = 13546 kg / m³

g is accelerqation due to gravity 9.8 m/s²

therefore h = 2*0.487*1/ 10 * 10^-3*13546*9.8 = +0.733 mm

therefore here as we have a positive error of  +0.733 mm due to capillary force

so air pressure is 760 + 0.733mm = 760.733 mm

error = 760 - 760.733  /760 = 0.964% error.