$A$ thin metal wire of density $\rho$ floats on the water surface horizontally. If it is $\text{NOT}$ to sink in water,then the maximum radius of the wire is proportional to $(T = \text{surface tension of water}, g = \text{gravitational acceleration})$.

Surface tension may be defined as

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 the force becomes smaller than the weight of the drop, the drop gets detached from the dropper.
$1.$ 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
$(A)$ $2 \pi r T$ $(B)$ $2 \pi R T$ $(C)$ $\frac{2 \pi r^2 T}{R}$ $(D)$ $\frac{2 \pi R^2 T}{r}$
$2.$ If $r=5 \times 10^{-4} \, m, \rho=10^3 \, kg \, m^{-3}, g=10 \, m/s^2, T=0.11 \, Nm^{-1}$, the radius of the drop when it detaches from the dropper is approximately
$(A)$ $1.4 \times 10^{-3} \, m$ $(B)$ $3.3 \times 10^{-3} \, m$
$(C)$ $2.0 \times 10^{-3} \, m$ $(D)$ $4.1 \times 10^{-3} \, m$
$3.$ After the drop detaches, its surface energy is
$(A)$ $1.4 \times 10^{-6} \, J$ $(B)$ $2.7 \times 10^{-6} \, J$
$(C)$ $5.4 \times 10^{-6} \, J$ $(D)$ $8.1 \times 10^{-6} \, J$
Give the answer for questions $1, 2$ and $3.$

Give a reason why some insects are able to walk on the water surface.

$A$ drop of some liquid of volume $0.04 \ cm^{3}$ is placed on the surface of a glass slide. Then another glass slide is placed on it in such a way that the liquid forms a thin layer of area $20 \ cm^{2}$ between the surfaces of the two slides. To separate the slides a force of $16 \times 10^{5} \ dyne$ has to be applied normal to the surfaces. The surface tension of the liquid is (in $dyne \ cm^{-1}$):

$A$ liquid drop of density $\rho$ is floating half immersed in a liquid of surface tension $S$ and density $\frac{\rho}{2}$. If the surface tension $S$ of the liquid is numerically equal to $10$ times the acceleration due to gravity $g$,then the diameter of the drop is: