Why is the surface tension of antiseptics kept low? Explain.

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 $3.$

$A$ 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 $(T = \text{surface tension of water}, g = \text{gravitational acceleration})$.

In most liquids, with a rise in temperature, the surface tension of the liquid

$A$ wire of length $10 \text{ cm}$ is gently placed horizontally on the surface of water having surface tension of $75 \times 10^{-3} \text{ N/m}$. What force is required to just pull up the wire from the water surface?

$A$ thin metal disc of radius $r$ floats on the water surface and bends the surface downwards along the perimeter,making an angle $\theta$ with the vertical edge of the disc. If the disc displaces a weight of water $W$ and the surface tension of water is $T$,then the weight of the metal disc is: