$A$ long cylindrical glass vessel has a pinhole of diameter $0.2 \,mm$ at its bottom. The depth to which the vessel can be lowered vertically in a deep water bath without the water entering into the vessel is (surface tension of water,$T=0.07 \,N/m$,acceleration due to gravity,$g=10 \,m/s^2$) (in $\,cm$)

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

Two soap bubbles combine to form a single bubble. In this process,the change in volume and surface area are respectively $V$ and $A$. If $P$ is the atmospheric pressure,and $T$ is the surface tension of the soap solution,the following relation is true :

An air bubble of radius $1 \ mm$ is at a depth of $8 \ cm$ below the free surface of a liquid column. If the surface tension and density of the liquid are $0.1 \ N \ m^{-1}$ and $2000 \ kg \ m^{-3}$ respectively, by what amount is the pressure inside the bubble greater than the atmospheric pressure (in $N \ m^{-2}$)? (Take $g = 10 \ m \ s^{-2}$)

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

$A$ soap bubble $(S.T = 30\, dyne/cm)$ has a radius of $1\, cm$. The work done in doubling its radius would be ........ $ergs$.