Surface tension

A glass capillary tube is of the shape of a truncated cone with an apex angle $\alpha$ so that its two ends have cross sections of different radii. When dipped in water vertically, water rises in it to a height $h$, where the radius of its cross section is $b$. If the surface tension of water is $S$, its density is $\rho$, and its contact angle with glass is $\theta$, the value of $h$ will be ($g$ is the acceleration due to gravity)

On heating water, bubbles being formed at the bottom of the vessel detatch and rise. Take the bubbles to be spheres of radius $R$ and making a circular contact of radius $r$ with the bottom of the vessel. If $r\ll R$, and the surface tension of water is $T$, value of $r$ just before bubbles detatch is :

(density of water is $\rho_w$)

An air bubble of radius 0.1 cm is in a liquid having surface tension 0.06 N/m and density $10^3$ kg/m$^3$. The pressure inside the bubble is 1100 Nm$^{-2}$ greater than the atmospheric pressure. At what depth is the bubble below the surface of the liquid ? ($g=9.8$ ms$^{-2}$)

Two soap bubbles coalesce to form a single bubble. If V is the subsequent change in volume of contained air and S the change in total surface area, T is the surface tension and P atmospheric pressure, which of the following relation is correct ?

A large number of liquid drops each of radius r coalesce to from a single drop of radius R. The energy released in the process is converted into kinetic energy of the big drop so formed. The speed of the big drop is (given surface tension of liquid T, density ρ)

Four point charges, each of $+q$, are rigidly fixed at the four corners of a square planar soap film of side '$a$'. The surface tension of the soap film is $\gamma$. The system of charges and planar film are in equilibrium, and $a=k\left[\dfrac{q^2}\gamma\right]^{1/N}$, where '$k$' is a constant. Then $N$ is

If the radius of the opening of the dropper is $r$, the vertical force due to the surface tension on the drop of radius $R$ (assuming $r\ll R$) is

If $r=5\times10^{-4}$ m, $\rho=10^3$ kg m⁻³, $g=10$ m s⁻², $T=0.11$ N m⁻¹, the radius of the drop when it detaches from the dropper is approximately

After the drop detaches, its surface energy is

Two soap bubbles A and B are kept in a closed chamber where the air is maintained at pressure 8 N/m². The radii of bubbles A and B are 2 cm and 4 cm, respectively. Surface tension of the soap-water used to make bubbles is 0.04 N/m. Find the ratio $n_B/n_A$, where $n_A$ and $n_B$ are the number of moles of air in bubbles A and B, respectively. [Neglect the effect of gravity.]

A glass tube of uniform internal radius ($r$) has a valve separating the two identical ends. Initially, the valve is in a tightly closed position. End 1 has a hemispherical soap bubble of radius $r$. End 2 has sub-hemispherical soap bubble as shown in figure. Just after opening the valve,

Water is filled up to a height $h$ in a beaker of radius $R$ as shown in the figure. The density of water is $\rho$, the surface tension of water is $T$ and the atmospheric pressure is $P_0$. Consider a vertical section ABCD of the water column through a diameter of the beaker. The force on water on one side of this section by water on the other side of this section has magnitude