If two soap bubbles of different radii are connected by a tube,

$A$ glass capillary tube of inner diameter $0.28 \ mm$ is lowered vertically into water in a vessel. The pressure to be applied on the water in the capillary tube so that the water level in the tube is the same as that in the vessel (in $N/m^2$) is:
Surface tension of water $= 0.07 \ N/m$
Atmospheric pressure $= 10^5 \ N/m^2$

Under isothermal conditions, two soap bubbles of radii $r_1$ and $r_2$ combine to form a single soap bubble of radius $R$. If $P$ is the outside pressure, find the surface tension $T$ of the soap solution.

$A$ drop of water with a volume of $0.05 \ cm^3$ is pressed between two glass plates,causing it to spread between the plates. The area of contact with each plate is $40 \ cm^2$. If the surface tension of water is $70 \ dyne/cm$,the minimum normal force required to separate the two glass plates in newtons is approximately: (assuming the angle of contact is zero)

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

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