$A$ 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):

$A$ circular film of a liquid has an area of $10 \,cm^2$. If the work done in increasing its radius to two times the initial radius is $8 \times 10^{-3} \,J$, the surface tension of the liquid is $\left(1+\frac{1}{\alpha}\right) \,N/m$. The value of $\alpha$ is

$A$ paper disc of radius $R$ from which a circular hole of radius $r$ is cut out,is floating in a liquid of surface tension $T$. The force on the disc due to surface tension will be

$A$ 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 $= 980 \ cm \ s^{-2}$]

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 $3.$

Water is filled up to a height $h$ in a beaker of radius $R$ as shown in the figure. The density of water is $\rho$,the surface tension of water is $T$ and the atmospheric pressure is $P_0$. Consider a vertical section $ABCD$ of the water column through a diameter of the beaker. The force on water on one side of this section by water on the other side of this section has magnitude