A capillary tube of radius 'r' is dipped vertically in a liquid of density '$ρ$' and surface tension 'T'. If the angle of contact is $heta$, the pressure difference between the two sides of the curved surface at the top of the meniscus is:

The excess pressure inside a spherical drop of radius $r$ of a liquid of surface tension T is

One end of a uniform glass capillary tube of radius$r = 0.025{\rm{cm}}$ is immersed vertically in water to a depth $h = 1{\rm{cm}}$. The excess pressure in ${\rm{N}}{{\rm{m}}^{ - 2}}$ required to blow an air bubble out of the tube (Surface tension of water = $7 imes {10^{ - 2}}{\rm{N}}{{\rm{m}}^{ - 1}}$, Density of water = ${10^3}{\rm{kg}}{{\rm{m}}^{ - 3}}$ , Acceleration due to gravity= 10${\rm{m}}{{\rm{s}}^{ - 2}}$

Water has been filled in a vertical capillary tube upto the top after closing its lower end by a finger Consider radius of capillary r = 1 mm, surface tension is 70 dyne/cm and g = 980 cm/sec . Which of the option is correct when finger is removed?

A spherical liquid drop of radius 'r' is divided into 'n' identical droplets. If surface tension is 'T', the work done in this process is:

Water rises to a height of 16.3 cm in a capillary of height 18 cm above the water level. If the tube is cut at a height of 12 cm in the capillary tube,

The excess pressure inside a spherical drop of radius $r$ of a liquid of surface tension T is

The excess pressure inside a liquid drop is given by:

Two spherical bubbles of radii ${r_1}$ and ${r_2}$ coalesce in vacuum under isothermal conditions. The radius $R$ of the resulting bubble is

Inside a liquid drop of radius $R$, the excess pressure is $P$. If two such drops coalesce to form a single larger drop, what will be the excess pressure inside the larger drop?

When a glass capillary tube of radius 0.015 cm is dipped in water, the water rises to height of 15 cm within it. Assuming contact angle between water and glass to be $0^\circ$, the surface tension of water is $[{{\rm{\rho }}_{{\rm{water}}}} = 1000{\rm{kg}}{{\rm{m}}^{ - 3}},{\rm{g}} = 9.81{\rm{m}}{{\rm{s}}^{ - 2}}]$