Small droplets of a liquid are usually more spherical in shape than larger drops of the same liquid because

$A$ rectangular film of liquid is extended from $(4 \, cm \times 2 \, cm)$ to $(5 \, cm \times 4 \, cm).$ If the work done is $3 \times 10^{-4} \, J,$ the value of the surface tension of the liquid is ............ $N \, m^{-1}$

The temperature at which the surface tension of water is zero is:

Consider a bowl filled with water on which some black pepper powder has been sprinkled uniformly. Now,a drop of liquid soap is added at the centre of the surface of water. The picture of the surface immediately after this will look like:

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 $3.$

$A$ metal sheet $4 \ m$ on a side and of negligible thickness is attached to a balance and inserted into a container of fluid. The balance to which the metal sheet is attached reads $0.50 \ N$ and the contact angle is found to be $0^{\circ}$. $A$ small amount of oil is then spread over the metal sheet. The contact angle now becomes $180^{\circ}$ and the balance now reads $0.49 \ N$. The surface tension of the fluid is: