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

If the excess pressure inside a soap bubble of radius $3 \,mm$ is equal to the pressure of a water column of height $0.8 \,cm$, then the surface tension of the soap solution is ( $\rho_{\text{water}} = 1000 \,kg/m^3, g = 9.8 \,m/s^2$ ).

The excess pressure inside a first spherical drop of water is three times that of a second spherical drop of water. Then the ratio of the mass of the first spherical drop to that of the second spherical drop is

The excess pressure inside a spherical drop of water is three times that of another drop of water. The ratio of their surface area is

The excess pressure inside a soap bubble is thrice the excess pressure inside a second soap bubble. The ratio between the volume of the first and the second bubble is:

In air,a charged soap bubble of radius $R$ breaks into $64$ small soap bubbles of equal radius $r$. The ratio of mechanical force per unit area of the big soap bubble to that of a small bubble is