Vertical Surface Tension Force on Spherical Dropper Drop

To find the vertical force due to the surface tension on the spherical drop formed at the opening of the dropper, we consider the boundary where the drop attaches to the dropper's opening:

1. Geometry of the Contact Boundary:
   The opening of the dropper is a circle of radius $r$. The spherical drop has a radius $R$ with $r \ll R$.
   The boundary of contact between the drop and the dropper is a circle of circumference:
   $L = 2\pi r$

2. Direction of Surface Tension Force:
   The surface tension force $T$ acts tangentially to the spherical surface along this circular boundary of contact.
   If $\theta$ is the angle made by the tangent to the drop surface with the horizontal at the circle of radius $r$, then from the spherical geometry:
   $\sin\theta = \frac{r}{R}$

3. Vertical Component of Surface Tension Force:
   The net vertical force $F_v$ pulling the drop upwards towards the dropper is given by the integral of the vertical component of the surface tension force along the entire circumference:
   $F_v = (2\pi r) \times T \sin\theta$

   Substituting $\sin\theta = \frac{r}{R}$:
   $F_v = 2\pi r T \left(\frac{r}{R}\right) = \frac{2\pi r^2 T}{R}$

(Alternatively, this can be obtained via the excess pressure method: The excess pressure inside the spherical drop is $\Delta P = \frac{2T}{R}$. The net upward force holding the liquid column at the opening of cross-sectional area $A = \pi r^2$ is $F = \Delta P \cdot A = \frac{2T}{R} \cdot (\pi r^2) = \frac{2\pi r^2 T}{R}$.)

Therefore, the vertical force due to surface tension is $\frac{2\pi r^2 T}{R}$, which corresponds to Option C.