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 . . . . . . .

$A$ drop of water with a volume of $0.05 \ cm^3$ is pressed between two glass plates,causing it to spread between the plates. The area of contact with each plate is $40 \ cm^2$. If the surface tension of water is $70 \ dyne/cm$,the minimum normal force required to separate the two glass plates in newtons is approximately: (assuming the angle of contact is zero)

Air is pushed into a soap bubble to increase its radius from $R$ to $2R$. In this case, the pressure inside the bubble

$A$ certain number of spherical liquid drops of radius $r$ coalesce to form a single drop of radius $R$ and volume $V$. If $T$ is the surface tension of the liquid,which one of the following statements is true for the energy $(E)$ in the process?

Two small drops of mercury each of radius $r$ coalesce to form a single large drop of radius $R$. The ratio of the total surface energies before and after the change is

When two soap bubbles of radius $r_1$ and $r_2$ $(r_2 > r_1)$ coalesce,the radius of curvature of the common surface is: