The surface tension of a soap solution is $25 \times 10^{-3} \, N/m$. The excess pressure inside a soap bubble of diameter $1 \, cm$ is ....... $Pa$.

Work done in splitting a drop of water of $1 \, mm$ radius into $10^6$ droplets is (Surface tension of water $= 72 \times 10^{-3} \, J/m^2$).

If two soap bubbles each of radius $2 \ cm$ combine in vacuum under isothermal conditions,then the radius of the new bubble formed is

Two soap bubbles combine to form a single bubble. In this process,the change in volume and surface area are respectively $V$ and $A$. If $P$ is the atmospheric pressure,and $T$ is the surface tension of the soap solution,the following relation is true :

$A$ large number of water drops,each of radius $r$,combine to form a single drop of radius $R$. If the surface tension is $T$ and the mechanical equivalent of heat is $J$,the rise in heat energy per unit volume will be:

The energy needed for breaking a liquid drop of radius $R$ into $216$ droplets, each of radius $r$, is $x$ times $TR^2$. The value of $x$ is [$T =$ surface tension of the liquid]. (in $\pi$)