# Measuring surface tension

#### boks

Joined Oct 10, 2008
218
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?

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#### mik3

Joined Feb 4, 2008
4,843
The perimeter of a rectangle is 2(l+w)

where
l=length
w=width

Are you trying to find the surface submerged into the liquid?
Is this surface includes the bottom side of the rod or just the side surface?

#### studiot

Joined Nov 9, 2007
4,998
I remember this one from my GCE practical exam in the 1960s

We had to float a needle on the surface of a dish of water , then drag the needle up the meniscus with a magnet, balancing the surface tension against the magnetic force to measure the tension.

#### boks

Joined Oct 10, 2008
218
The problem is stated in its original form. I forgot to add an illustration, now it's done.

I'm trying to find the surface tension. I'm familiar with problems of this kind where the submerged item is a Wilhelmy plate. Then P = 2(l + t) where l is the length and t is the thickness. This is the perimeter of the cross section between the plate and the liquid. The cross section isn't that obvious - at least not to me - when the plate is replaced by a rod.

Assuming P = 2(1cm + 0.2 cm) = 2.4 cm, I get

$$\gamma = \frac{0.49 mN}{2.4 cm \cdot cos(0)}=20.4 mN/m$$

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#### mik3

Joined Feb 4, 2008
4,843
If it is the perimeter of the cross section, ie around the plate, then for the rod the perimeter is 2*pi*R.