$A$ soap bubble has radius $R$ and thickness $d$ $(d \ll R)$ as shown. It collapses into a spherical drop. The ratio of excess pressure in the drop to the excess pressure inside the bubble is

Due to surface tension, the excess pressure inside a smaller drop is $9$ units. If $27$ smaller drops of the same liquid combine, then the excess pressure inside the bigger drop is (in $units$)

The excess of pressure,due to surface tension,on a spherical liquid drop of radius $R$ is proportional to

In Jager's method,at the time of bursting of the bubble,

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 ( $\rho_{\text{water}} = 1000 \,kg/m^3, g = 9.8 \,m/s^2$ ).

$A$ soap bubble,having a radius of $1\; mm$,is blown from a detergent solution having a surface tension of $2.5 \times 10^{-2}\; N/m$. The pressure inside the bubble is equal to the pressure at a point $Z_{0}$ below the free surface of water in a container. Taking $g=10\; m/s^{2}$ and the density of water $\rho = 10^{3}\; kg/m^{3}$,the value of $Z_{0}$ is......$cm$.