On mixing salt in water,the surface tension of water will:

The unit of surface tension in the $SI$ system is

$A$ small spherical droplet of density $d$ is floating exactly half immersed in a liquid of density $\rho$ and surface tension $T$. The radius of the droplet is (take note that the surface tension applies an upward force on the droplet).

$A$ pin or a needle floats on the surface of water,the reason for this is:

$A$ rectangular film of liquid is expanded from $(5 \text{ cm} \times 4 \text{ cm})$ to $(7 \text{ cm} \times 8 \text{ cm})$. If the work done is $3 \times 10^{-4} \text{ J}$,the surface tension of the liquid is (nearly):

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 $3.$