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:

$A$ metal wire of density $\rho$ floats on the water surface horizontally. If it is not to sink in water,then the maximum radius of the wire is ($T$ = surface tension of water,$g$ = gravitational acceleration).

$A$ thin liquid film formed between a $U$-shaped wire and a light slider supports a weight of $1.5 \times 10^{-2} \; N$ (see figure). The length of the slider is $30 \; cm$ and its weight is negligible. The surface tension of the liquid film is (in $; N m^{-1}$)

The work done in increasing the size of a soap film from $10 \, cm \times 6 \, cm$ to $10 \, cm \times 11 \, cm$ is $3 \times 10^{-4} \, J$. The surface tension of the film is:

Consider a spherical drop of radius $R$. The surface tension of the liquid is $\sigma$. The force of surface tension on the shaded sub-hemisphere due to the remaining drop is $0.5\,\pi \sigma R$. The value of the angle $\theta$ (angle subtended by the sub-hemisphere at the center of the drop) is ......$^o$.

$A$ $U$-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?