If the work done in increasing the size of a soap film from $10\, cm \times 6\, cm$ to $60\, cm \times 11\, cm$ is $2 \times 10^{-4}\, J$,what is the surface tension?

$A$ $U$-shaped wire is dipped in a soap solution and removed. The thin soap film formed between the wire and the light slider supports a weight of $1.5 \times 10^{-2} \; N$ (which includes the small weight of the slider). The length of the slider is $30 \; cm$. What is the surface tension of the film?

$A$ needle is $7 \ cm$ long. Assuming that the needle is not wetted by water,what is the weight of the needle,so that it floats on water (in $g \ wt$)? $\left[T = \text{surface tension of water} = 70 \ dyne/cm\right]$ [acceleration due to gravity $= 980 \ cm \ s^{-2}$]

On heating water,bubbles formed at the bottom of the vessel detach and rise. Assume the bubbles are spheres of radius $R$ and make a circular contact of radius $r$ with the bottom of the vessel. If $r << R$ and the surface tension of water is $T$,find the value of $r$ just before the bubbles detach. (Density of water is $\rho_{w}$)

$A$ soap film is formed on a circular frame. $A$ loop of thread is lying on the film. If the film inside the loop is broken,then the tension in the thread will be: ($R =$ radius of the loop,$T =$ surface tension of the liquid)

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 $3.$