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 . . . . . . .

Consider a water droplet of diameter $0.2 \,mm$ where the outside pressure is $1.5 \,N / cm^2$ at $25^{\circ} C$. The pressure inside the droplet, when the surface tension at $25^{\circ} C$ is $0.08 \,N / m$ is

Pressure inside a soap bubble is greater than the pressure outside by an amount: (given: $R =$ Radius of bubble,$S =$ Surface tension of bubble)

$A$ spherical soap bubble of radius $3\,cm$ is formed inside another spherical soap bubble of radius $6\,cm$. If the internal pressure of the smaller bubble of radius $3\,cm$ in the above system is equal to the internal pressure of another single soap bubble of radius $r\,cm$,then the value of $r$ is:

The work done in splitting a water drop of radius $R$ into $64$ droplets is ($T=$ Surface tension of water). (in $\pi TR^2$)

$A$ water drop of $0.01 \ cm^3$ is squeezed between two glass plates and spreads into an area of $10 \ cm^2$. If the surface tension of water is $70 \ dyne/cm$,then the normal force required to separate the glass plates from each other will be: (in $N$)