Physics

Problem

When liquid medicine of density

\rho

is to be put in the eye, it is done with the help of a dropper. As the bulb on the top of the dropper is pressed, a drop forms at the opening of the dropper. We wish to estimate the size of the drop. We first assume that the drop formed at the opening is spherical because that requires a minimum increase in its surface energy. To determine the size, we calculate the net vertical force due to the surface tension

T

when the radius of the drop is

R

. When this force becomes smaller than the weight of the drop, the drop gets detached from the dropper. 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 \ll 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}

Intuitive explanation

The key insight is that surface tension acts along the perimeter where the drop connects to the dropper, pulling upward with a force proportional to that contact line. The drop detaches when gravity overcomes this upward pull—so you need to find the vertical component of the surface tension force acting around the circular opening of radius $r$ . Since $r \ll R$ , the contact line is essentially a circle of radius $r$ , and surface tension $T$ acts tangentially along it; the vertical component depends on the contact angle at the dropper opening.

Intuitive explanation

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