What force are we looking for?

The liquid drop is attached to the circular opening of the dropper of radius $r$.

Surface tension $T$ acts tangentially along the contact line all around the rim.

So, the total force due to surface tension is obtained by: $F = T \times (\text{perimeter of contact}) \times (\text{vertical component factor})$

Perimeter of the contact line

The contact line is the circular rim of the dropper opening, so its perimeter is $2\pi r$

Vertical component of surface tension

Since the drop is spherical with radius $R$, and the opening radius is much smaller than the drop radius ($r \ll R$), the surface near the opening makes a small angle $\theta$ with the horizontal.

From geometry of the sphere, $\sin\theta \approx \frac{r}{R}$ for small $r$.

The vertical component of surface tension per unit length is therefore $T\sin\theta = T\frac{r}{R}$

Total vertical force

Multiply by the full circumference: $F = (2\pi r)\left(T\frac{r}{R}\right)$

Hence, $F = \frac{2\pi r^2 T}{R}$

Matching with options

This corresponds to: $\boxed{\frac{2\pi r^2 T}{R}}$

So the correct option is C.

Comparison with stored answer

Stored correct answer: C

Our derived answer: C

Therefore, they agree.