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?

$A$ wire of length $10 \text{ cm}$ is gently placed horizontally on the surface of water having surface tension of $75 \times 10^{-3} \text{ N/m}$. What force is required to just pull up the wire from the water surface?

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

$A$ 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):

$A$ $U$-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?

Figure $(a)$ shows a thin liquid film supporting a small weight $= 4.5 \times 10^{-2} \, N$. What is the weight supported by a film of the same liquid at the same temperature in Figure $(b)$ and $(c)$? Explain your answer physically.