The excess pressure inside a bubble in water is $P_1$. The excess pressure inside a drop of the same radius is $P_2$. Then:

The surface tension of a soap solution is $2 \times 10^{-2} \ N/m$. To blow a bubble of radius $1 \ cm$,the work done is:

Two soap bubbles of radii $2 \, cm$ and $4 \, cm$ join to form a double bubble in air. The radius of curvature of the interface is .......... $cm$.

$1000$ small water drops,each of radius $r$ and charge $q$,coalesce together to form one large spherical drop. The potential of the big drop is larger than that of the smaller drop by a factor of:

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 $....Pa \left( g = 10\,m\,s^{-2} \right)$

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.