Q.There is a small bubble at one end and bigger bubble at other end of a pipe. Which among the following will happen ?

Excess pressure = 4T/r, so pressures are in the inverse ratio of radii: 1:2. Answer (D).

For a soap bubble (two surfaces), the excess pressure inside is P = 4T/r, inversely proportional to the radius.

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The excess pressure inside a soap bubble of radius r is 4T/r.

For a single liquid drop (one surface), the excess pressure inside is $\Delta P = \dfrac{2T}{r}$, from the balance between surface tension force and the pressure difference across the curved surface.

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Because excess pressure in a bubble varies as $1/r$, the ratio of excess pressures is the inverse of the ratio of radii.

For a soap bubble (which has two liquid surfaces, inner and outer), the excess pressure inside is:

$\Delta P = 4T/r$

where $T$ is the surface tension (same for both bubbles, same soap solution) and $r$ is the radius.

Given $r_1 : r_2 = 3 : 2$:

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A soap bubble in air has two surfaces (inside and outside the thin film), so its excess pressure is double the single-surface case: 4S/R.

For a spherical liquid drop, which has only one free surface, surface tension gives an excess pressure inside of

P = 2S/R

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Excess pressure inside a drop of diameter $d$ is $4T/d$.

For a spherical liquid drop (one surface), the excess pressure inside due to surface tension is $\Delta P=\dfrac{2T}{r}$, where $r$ is the drop's ra …

The excess pressure inside a soap bubble is P = 4T/r, so it is inversely proportional to radius; a bigger bubble has lower excess pressure.

A soap bubble has two liquid surfaces (inner and outer), so its excess pressure is:

$P = \frac{4T}{r}$

where T is the surface tension of the soap solution and r is the bubble's radius. For two bubbles of radii $r_1$ and $r_2$:

$\frac{P_1}{P_2} = \frac{r_2}{r_1}$

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Pressure inside a soap bubble is greater than atmospheric.

A soap bubble has two liquid surfaces (inner and outer). Surface tension curves these surfaces, and the excess pressure inside a soap bubble is

ΔP = 4T/r,

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The pressure of the air inside a bubble is greater than outside (it increases) due to surface tension.

A soap bubble has two surfaces, so the excess pressure inside over the outside is $\Delta P = \dfrac{4T}{r}$ (for a liquid drop or air bubble in liquid it is 2T/r).

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The ratio of the two bubble volumes is 8 : 1.

For a soap bubble the excess pressure over the atmosphere is $\Delta P = \dfrac{4T}{r}$, so $r = \dfrac{4T}{\Delta P}$.

Taking atmospheric pressure as 1.00 atm:

• Bubble 1: ΔP₁ = 1.01 − 1.00 = 0.01 atm
• Bubble 2: ΔP₂ = 1.02 − 1.00 = 0.02 atm …

Because excess pressure inside a bubble is P = 4T/r, the smaller bubble (smaller r) has higher internal pressure, so air flows from the smaller bubble into the bigger one -- the bigger bubble grows while the smaller one shrinks and collapses.

For a soap bubble (which has two liquid surfaces, an inner and an outer film) of radius r and surface tension T, the excess pressure inside the bubble over the outside atmospheric pressure is

P_excess = 4T/r

This shows that the excess pressure is INVERSELY proportional to the radius: a smaller bubble has a smaller r, and hence a LARGER excess pressure than a bigger bubble of the same liquid.

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