To meaure the surface tension of a liquid, a rod of r = 1 mm radius and l = 1 cm length is dipped horizontally into the liquid (see figure). The force required to pull the rod out of the completely wetting liquid is measured to be 0,49 mN. What is the surface tension of the liquid?
\(
\gamma = \frac{F}{P \cdot cos(\theta)}
\)
This formula can be applied to the Wilhemly Plate Technique for measuring surface tension. When a rectangular plate having length l and thickness t is submerged, P, the perimeter of the plate, is 2(l + t). When ithe submerged item is a rod, is it reasonable to assume P = 2(2r + l), i.e. the liquid surrounds half the surface area while the rod is being pulled out?
\(
\gamma = \frac{F}{P \cdot cos(\theta)}
\)
This formula can be applied to the Wilhemly Plate Technique for measuring surface tension. When a rectangular plate having length l and thickness t is submerged, P, the perimeter of the plate, is 2(l + t). When ithe submerged item is a rod, is it reasonable to assume P = 2(2r + l), i.e. the liquid surrounds half the surface area while the rod is being pulled out?
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