Three small identical bubbles of water having the same charge on each coalesce to form a bigger bubble. Then the ratio of the potentials on one initial bubble and that on the resultant bigger bubble is:

There is an air bubble of radius $1.0\,mm$ in a liquid of surface tension $0.075\,N\,m^{-1}$ and density $1000\,kg\,m^{-3}$ at a depth of $10\,cm$ below the free surface. The amount by which the pressure inside the bubble is greater than the atmospheric pressure is $....Pa \left( g = 10\,m\,s^{-2} \right)$

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.

Two small drops of mercury,each of radius $R$,coalesce to form a single large drop. The ratio of the total surface energies before and after the change is

Under isothermal conditions, two soap bubbles of radii $a$ and $b$ coalesce to form a single bubble of radius $c$. If the external pressure is $P$, then the surface tension of the bubbles is:

The work done in splitting a water drop of radius $R$ into $64$ droplets is ($T=$ Surface tension of water). (in $\pi TR^2$)