Question

Two soap bubbles of radii 4 cm and 3 cm respectively coalesce under isothermal conditions to form a single bubble. What is the radius of the new single bubble?

• A. $3cm$
• B. $4cm$
• C. $5cm$
• D. $6cm$

Answer 1

The correct option is C $5cm$

Given:

Radius of first soap bubble, $r_1=4cm$

Radius of second soap bubble, $r_2=3cm$

Let,

$P_1=\frac{4T}{(r_1{)}^2},V_1=\frac{4}{3}\pi (r_1{)}^3$ be the excess pressure inside first soap bubble and volume of first soap bubble respectively.

$P_2=\frac{4T}{(r_2{)}^2},V_2=\frac{4}{3}\pi (r_2{)}^3$ be the excess pressure inside second soap bubble and volume of second soap bubble respectively.

And $P=\frac{4T}{r^2},V_2=\frac{4}{3}\pi r^3$be the excess pressure inside new soap bubble, volume and radius of new soap bubble respectively.

The two bubbles combine isothermally hence,

$PV=P_1{V}_1+P_2{V}_2$

$\frac{4T}{r}\frac{4}{3}\pi r^3=\frac{4T}{\left(r_1\right)}\frac{4}{3}\pi (r_1{)}^3+\frac{4T}{\left(r_2\right)}\frac{4}{3}\pi (r_2{)}^3$

$r^2=(r_1{)}^2+(r_2{)}^2$

$r=\sqrt{(}(r_1{)}^2+(r_2{)}^2)$

$r=\sqrt{(}(4{)}^2+(3{)}^2)$

$r=\sqrt{2}5$

$r=5cm$