When liquid medicine of density `rho` is to be put in the eye, it is done with the help of a dropper. As the bylb on the top of the dropper is pressed. A drop frons 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=0.11 when the radius of the drop is R=1.4mm. when this force become smaller than the weight of the drop the drop gets detached from the dropper.
After the drop detaches, its surface energy is

Options

A. `1.4 xx 10^(-6) J`
B. `3.3 xx 10^(-6) J`
C. `5.4 xx 10^(-6) J`
D. `8.1 xx 10^(-6) J`

Answer

B

Explanation

Surface energy of the drop is equal to surface tension multiplied with its surface area. `U=T(4pi R^(2))`. On substituting values from the previous question we get `U=2.7 xx 10^(-6)J`, hence option (b) is correct.