Two soap bubbles of radii r1 and r2 and pressure P1 and P2 respectively are stuck together with an intermediate film separating them- Choose the correct options-A- The radius of curvature of this film is r 1 r 2- r 1- r 2-B- The radius of curvature of this film is r1 r22- r 1- r 2-C- If r1-r2 and P2-P1- the intermediate film will bulge towards the center of radius r1D- If r1-r2 and P2-P1- the intermediate film will bulge towards the center of radius r2

The correct options are A The radius of curvature of this film is r_1r_2/|r_1 r_2| C If r_1 gt; r_2 and P_2 gt; P_1, the intermediate film will bulge towa ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Physics

Surface Energy

Two soap bubb...

Question

Two soap bubbles of radii $r_{1}$ and $r_{2}$ and pressure $P_{1}$ and $P_{2}$ respectively are stuck together with an intermediate film separating them. Choose the correct option(s).

The radius of curvature of this film is

\frac{r_{1} r_{2}}{| r_{1} - r_{2} |}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

The radius of curvature of this film is

\frac{r_{1} r_{2}}{| r_{1} + r_{2} |}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

r_{1} > r_{2}

and

P_{2} > P_{1}

, the intermediate film will bulge towards the center of radius

r_{1}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

r_{1} > r_{2}

and

P_{2} > P_{1}

, the intermediate film will bulge towards the center of radius

r_{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct options are
A The radius of curvature of this film is $\frac{r_{1} r_{2}}{| r_{1} - r_{2} |}$
C If $r_{1} > r_{2}$ and $P_{2} > P_{1}$ , the intermediate film will bulge towards the center of radius $r_{1}$
Excess pressure inside a soap bubble is $\frac{4 T}{r}$ where $T$ is the surface tension.
If $P$ is the atmospheric pressure and if pressure inside soap bubble are $P_{1}$ and $P_{2}$ , then
$P_{1} = \frac{4 T}{r_{1}} + P, P_{2} = \frac{4 T}{r_{2}} + P$
Now, difference of pressure between these two films will decide on which side intermediate film will bulge and radius of curvature of this film.
$∴ P_{1} - P_{2} = 4 T | (\frac{1}{r_{1}} - \frac{1}{r_{2}}) | = \frac{4 T}{r_{12}}$
$\Rightarrow \frac{1}{r_{12}} = | \frac{1}{r_{1}} - \frac{1}{r_{2}} |$
$\Rightarrow r_{12} = \frac{r_{1} r_{2}}{| r_{1} - r_{2} |}$
As pressure is more on concave side, therefore it will bulge on that side where pressure is less.
Hence, if $r_{1} > r_{2}$ and $P_{2} > P_{1}$ , the intermediate film will bulge towards the center of radius $r_{1}$ .

Suggest Corrections

Similar questions

When two soap bubbles of radius $r_{1}$ and $r_{2}$ ( $r_{2}$ > $r_{1}$ ) coalesce, the radius of curvature of common surface is

When resistors $R_{1} a n d R_{2}$ are connected in series across a battery of potential difference V, the power dissipated by the resistors are $P_{1} a n d P_{2}$ respectively. When they are connected in parallel across a battery of potential difference E, the power dissipated is $P_{3} a n d P_{4}$ respectively. Which one of the following is true?