A soap bubble of radius r₁ is placed on another soap bubble of radius r₂(r₁

Correct Answer is : r1r2r2−r1{r₁ r₂}{r₂-r₁}r2−r1r1r2

Correct Answer is : r1r2r2−r1{r₁ r₂}{r₂-r₁}r2−r1r1r2

Test Index

VITEEE 2021 Paper

Section: Physics

Question : 4 of 80

Marks: +1, -0

A soap bubble of radius

r_{1}

r_{2}(r_{1} < r_{2})

R

of the soapy film separating the two bubbles is

Solution:

Excess pressure in a soap bubble of radius

r_{1}=\frac{4 T}{r_{1}}

Excess pressure in a soap bubble of radius

r_{2}=\frac{4 T}{r_{2}}

The excess pressure of a soapy film separating the two bubbles with

R

as the radius of common surface

=\frac{4 T}{R}

Now

\frac{4 T}{R}=\frac{4 T}{r_{1}}-\frac{4 T}{r_{2}}

\frac{1}{R}=\frac{1}{r_{1}}-\frac{1}{r_{2}}

\frac{1}{R}=\frac{r_{2}-r_{1}}{r_{1} r_{2}}

R=\frac{r_{1} r_{2}}{r_{2}-r_{1}}

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