If the work done in blowing a bubble of volume $V$ is $W$,then the work done in blowing the bubble of volume $2V$ from the same soap solution will be

Fill in the blanks:
$(i)$ $A$ bubble in water has .......... free surface.
$(ii)$ $A$ bubble in air has .......... free surface.
$(iii)$ $A$ rain drop has .......... free surface.

What is the pressure inside the drop of mercury of radius $3.00 \; mm$ at room temperature? Surface tension of mercury at that temperature $(20 \; ^{\circ}C)$ is $4.65 \times 10^{-1} \; N m^{-1}$. The atmospheric pressure is $1.01 \times 10^{5} \; Pa$. Also,calculate the excess pressure inside the drop.

Due to surface tension, the excess pressure inside a smaller drop is $9$ units. If $27$ smaller drops of the same liquid combine, then the excess pressure inside the bigger drop is (in $units$)

$A$ container,whose bottom has round holes with diameter $0.1 \ mm$ is filled with water. The maximum height in $cm$ up to which water can be filled without leakage will be ........ $cm$. (Surface tension $= 75 \times 10^{-3} \ N/m$ and $g = 10 \ m/s^2$)

Small drops of liquid of the same radius coalesce to form a big drop. The ratio of the total surface energies after and before the change is: