Two spherical soap bubbles of radii $r_1$ and $r_2$ in vacuum combine under isothermal conditions. The resulting bubble has a radius equal to:

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. The radius of the bigger drop is $R$ and that of the smaller drop is $r$. Then the value of $n$ is:

If the excess pressure inside a soap bubble is balanced by an oil column of height $2\, mm$,then the surface tension of the soap solution will be $(r = 1\, cm$ and density $d = 0.8\, g/cm^3)$.

$A$ vessel having a small hole in the bottom must hold water without leakage when water is poured to a height of $7 \text{ cm}$. What is the radius of the hole (in $\text{ mm}$)? [Surface tension of water is $0.07 \text{ N/m}$, angle of contact is $0^{\circ}$, and $g = 10 \text{ m/s}^2$]

$A$ 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.

$A$ glass tube of uniform internal radius has a valve separating two identical ends. Initially,the valve is in a tightly closed position. End $1$ has a hemispherical soap bubble of radius $r$. End $2$ has a soap bubble with a radius of curvature $R$ $(R > r)$ as shown in the figure. Just after opening the valve,