Surface tension and the volume of drops

[madhuman]

Homework Statement

For a capillary, or a thin glass rod, the surface tension of a drop on it's end is given by T=kmg , where k is some constant which depends on the radius of the capillary or rod, and mg is the weight of the drop.
After experimentally finding values for the volume of the drops formed at the end of some rods, plot the volume of the drops against the radius of the rods used.
If there is a linear portion in the graph, find the value of k from it.

Homework Equations

T=kmg

The Attempt at a Solution

I got a graph that is linear, passing (approximately) through the origin. Now, I know that mass=volume*density, therefore volume is proportional to the weight of the drop, and hence proportional to the surface tension (from T=kmg). But I cannot understand how to get the value of k from the graph. All I understand about the graph is that it should pass through the origin (because when the radius of the rod =0, the volume of the drop is obviously 0).


Any help, please?

 

[anathema]

Thank you for your post. It seems like you have made good progress in your experiment and have a graph that shows a linear relationship between the volume of the drops and the radius of the rods. This is a great start!

To find the value of k from the graph, you can use the slope of the line. The slope represents the change in volume divided by the change in radius. In other words, it tells you how much the volume changes for every unit increase in radius.

Since we know that the volume is proportional to the weight of the drop, and the weight is proportional to the surface tension, we can say that the slope is also proportional to the surface tension. Therefore, you can find the value of k by calculating the slope of the line and multiplying it by the radius.

I hope this helps. Keep up the good work in your experiment!