The surface tension of a liquid at its boiling point

$A$ needle is $7 \ cm$ long. Assuming that the needle is not wetted by water,what is the weight of the needle,so that it floats on water (in $g \ wt$)? $\left[T = \text{surface tension of water} = 70 \ dyne/cm\right]$ [acceleration due to gravity $= 980 \ cm \ s^{-2}$]

In most liquids,with a rise in temperature,the surface tension of a liquid

$A$ soap film is formed on a circular frame. $A$ loop of thread is lying on the film. If the film inside the loop is broken,then the tension in the thread will be: ($R =$ radius of the loop,$T =$ surface tension of the liquid)

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

$n$ 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 [$T = \text{surface tension of liquid}, \rho = \text{density of liquid}$.]