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

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:

$A$ beaker of radius $15\, cm$ is filled with a liquid of surface tension $0.075\, N/m$. The force across an imaginary diameter on the surface of the liquid is:

$A$ 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

The force required to take away a flat circular plate of radius $2 \ cm$ from the surface of water is $[$ Surface tension of water $= 70 \times 10^{-3} \ Nm^{-1}, \pi = \frac{22}{7} ]$

$A$ small spherical droplet of density $d$ is floating exactly half immersed in a liquid of density $\rho$ and surface tension $T$. The radius of the droplet is (take note that the surface tension applies an upward force on the droplet).