Let $R_1$ and $R_2$ be the radii of two mercury drops. $A$ big mercury drop is formed from them under isothermal conditions. The radius of the resultant drop is

$A$ drop of water of volume $0.05\, cm^3$ is pressed between two glass plates,as a consequence of which it spreads and occupies an area of $40\, cm^2$. If the surface tension of water is $70\, dyne/cm$,then the normal force required to separate out the two glass plates will be in Newton.

When a big drop of water is formed from $n$ small drops of water, the energy loss is $3E$, where $E$ is the energy of the bigger drop. If $R$ is the radius of the bigger drop and $r$ is the radius of the smaller drop, then the number of smaller drops $(n)$ is:

The excess pressure inside a soap bubble of radius $2 \ cm$ is $50 \ dyne/cm^2$. The surface tension is

If the surface tension of a soap solution is $0.03 \, N/m$,then the excess pressure inside a soap bubble of diameter $6 \, mm$ over the atmospheric pressure will be:

Under isothermal conditions, two soap bubbles of radii $r_1$ and $r_2$ combine to form a single soap bubble of radius $R$. If $P$ is the outside pressure, find the surface tension $T$ of the soap solution.