Question

If the radius of the opening of the dropper is $r$, the vertical force due to the surface tension on the drop of radius $R$ (assuming $r<<R$) is:

• A. $2\pi rT$
• B. $2\pi RT$
• C. $\frac{2\pi r^{2}T}{R}$
• D. $\frac{2\pi R^{2}T}{r}$

Answer 1

The correct option is C $\frac{2\pi r^{2}T}{R}$

From the above figure we find that the component of surface tension will balance the weight of the drop.

$F_v=F\sin \theta$ where$F_v$ is the vertical component of the surface tension

So, $F_v=2\pi rT\sin \theta$

And $\sin \theta =\frac{r}{R}$

hence,$F_v=\frac{2\pi r^{2}T}{R}$ and Option C is correct.