If the surface area of the soap solution is increased,then its surface tension

$n$ number of liquid drops each of radius $r$ coalesce to form a single drop of radius $R$. The energy released in the process is converted into the kinetic energy of the big drop so formed. The speed of the big drop is [$T = \text{surface tension of liquid}, \rho = \text{density of liquid}$.]

$A$ thin metal disc of radius $r$ floats on the water surface and bends the surface downwards along the perimeter,making an angle $\theta$ with the vertical edge of the disc. If the disc displaces a weight of water $W$ and the surface tension of water is $T$,then the weight of the metal disc is:

When water is filled carefully in a glass,one can fill it to a height $h$ above the rim of the glass due to the surface tension of water. To calculate $h$ just before water starts flowing,model the shape of the water above the rim as a disc of thickness $h$ having semicircular edges,as shown schematically in the figure. When the pressure of water at the bottom of this disc exceeds what can be withstood due to the surface tension,the water surface breaks near the rim and water starts flowing from there. If the density of water,its surface tension and the acceleration due to gravity are $10^3 \ kg \ m^{-3}$,$0.07 \ N \ m^{-1}$ and $10 \ m \ s^{-2}$,respectively,the value of $h$ (in $mm$) is:

Work of $3.0 \times 10^{-4} \, J$ is required to be done in increasing the size of a soap film from $10 \, cm \times 6 \, cm$ to $10 \, cm \times 11 \, cm$. The surface tension of the film is

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