When liquid medicine of density $$\rho$$ is to be put in theeye, it is done with the help of a dropper. As the bulb onthe topof the dropperis pressed, a drop forms at the opening of the dropper. We wishto estimate the size of the drop. Wefirst 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 $$r = 5 \times 10^{-4} m, p = 10^3 kgm^{-3}, g = 10 ms^{-2}, T = 0.11 Nm^{-1}$$, the radius of the drop when it detaches from the dropper is approximately
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