$A$ liquid drop at temperature $t$,isolated from its surroundings,breaks into a number of droplets. The temperature of the droplets will be

The area of a liquid film is $60\, cm^2$ and its surface tension is $T = 20\, dyne/cm$. What is the work done to increase its area to $120\, cm^2$ (in $, erg$)?

$A$ soap bubble of radius $R$ is blown. After heating the solution, a second bubble of radius $2R$ is blown. The work required to blow the second bubble in comparison to that required for the first bubble is

If $1000$ droplets of water of surface tension $0.07\,N/m$,each having the same radius $1\,mm$,combine to form a single drop,the released surface energy in the process is: (Take $\pi = \frac{22}{7}$)

The work done in blowing a soap bubble of diameter $3 \ cm$ is (Surface tension of soap solution $= 0.035 \ N/m$). (in $\mu J$)

Fill in the blanks:
$(i)$ If the molecules of a body repel each other,then the surface area ...... and the surface energy ...... (choose between 'increases' or 'decreases').
$(ii)$ If the molecules of a body attract each other,then the surface area ...... and the surface energy ...... (choose between 'increases' or 'decreases').