The rain drops are in spherical shape due to

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 $3.$

$A$ square frame of each side $L$ is dipped in a soap solution and taken out. The force acting on the film formed is ($T =$ surface tension of soap solution). (in $TL$)

Consider a spherical drop of radius $R$. The surface tension of the liquid is $\sigma$. The force of surface tension on the shaded sub-hemisphere due to the remaining drop is $0.5\,\pi \sigma R$. The value of the angle $\theta$ (angle subtended by the sub-hemisphere at the center of the drop) is ......$^o$.

$A$ 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 $dyne \ cm^{-1}$):

$A$ wooden stick $2 \, m$ long is floating on the surface of water. The surface tension of water is $0.07 \, N/m$. By putting soap solution on one side of the stick,the surface tension is reduced to $0.06 \, N/m$. The net force on the stick will be ....... $N$.