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:

$A$ cylinder with a movable piston contains air under a pressure $p_1$ and a soap bubble of radius $r$. The pressure $p_2$ to which the air should be compressed by slowly pushing the piston into the cylinder for the soap bubble to reduce its size by half will be: (The surface tension is $\sigma$, and the temperature $T$ is maintained constant)

The diameter of a rain-drop is $0.02 \, cm$. If the surface tension of water is $72 \times 10^{-3} \, N/m$,then the pressure difference between the internal and external surfaces of the drop will be:

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 soap bubbles of radius $2 \ cm$ and $4 \ cm$,respectively,are in contact with each other. The radius of curvature of the common surface,in $cm$,is . . . . . . .

If two soap bubbles of different radii are connected by a tube,then