A vertical glass capillary tube of radius $r$ open at both ends contains some water (surface tension $T$ and density $\rho$ ). If $L$ be the length of the water column, then:

Excess pressure inside a soap bubble is three times that of the other bubble, then the ratio of their volumes will be

A cylinder with a movable piston contains air under a pressure $p_1$ and a soap bubble of radius $'r'$ . The pressure $p_2$ to which the air should be compressed by slowly pushing the piston into the cylinder for the soap bubble to reduce its size by half will be: (The surface tension is $\sigma $ , and the temperature $T$ is maintained constant)

If the excess pressure inside a soap bubble is balanced by oil column of height $2\; mm$, then the surface tension of soap solution will be ($r = 1 \,cm$ and density $d = 0.8\, gm/cc$)

Two narrow bores of diameter $5.0\, {mm}$ and $8.0\, {mm}$ are joined together to form a $U-$shaped tube open at both ends. If this ${U}$-tube contains water, what is the difference in the level of two limbs of the tube.

[Take surface tension of water ${T}=7.3 \times 10^{-2} \, {Nm}^{-1}$, angle of contact $=0, {g}=10\, {ms}^{-2}$ and density of water $\left.=1.0 \times 10^{3} \,{kg} \,{m}^{-3}\right]$ (in $mm$)

The excess of pressure inside a soap bubble is twice the excess pressure inside a second soap bubble. The volume of the first bubble is $n$ times the volume of the second where $n$ is