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

$A$ capillary type tube $AB$ is connected to a manometer $M$ as shown in the figure. Stopper $S$ controls the flow of air. $AB$ is dipped into a soap solution where surface tension is $T$. On opening the stopper for a while,a bubble is formed at $B$ end of the manometer and the level difference in manometer limbs is $h$. If $\rho$ is the density of the soap solution and $r$ is the radius of curvature of the bubble,then the surface tension $T$ of the liquid is given by:

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

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 excess pressure inside a soap bubble is three times that of another soap bubble. The ratio of their volumes is:

$A$ water drop is divided into $8$ equal droplets. The pressure difference between the inner and outer side of the big drop will be