$A$ 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

$A$ metal sheet $4 \ m$ on a side and of negligible thickness is attached to a balance and inserted into a container of fluid. The balance to which the metal sheet is attached reads $0.50 \ N$ and the contact angle is found to be $0^{\circ}$. $A$ small amount of oil is then spread over the metal sheet. The contact angle now becomes $180^{\circ}$ and the balance now reads $0.49 \ N$. The surface tension of the fluid is:

With an increase in temperature,surface tension of liquid (except molten copper and cadmium)

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:

$A$ 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?

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 ......$^o$.