Question

A soap bubble in vacuum has a radius of 3 cm ad another soap bubble in vacuum has a radius of 4 cm. if the two bubbles coalesce under isothermal condition, then the radius of the new bubble is

• A. 2.3 cm
• B. 4.5 cm
• C. 5 cm
• D. 7 cm

Answer 1

Answer: (C)

5 cm

Pressure inside the bubble in vacuum

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

Volume of bubble, $V=\frac{4}{3}\pi r^3$

As per question

$V=\frac{4}{3}\pi r^3$

Isothermal process obeys Boyle's law.

Hence $PV=constant$

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

$\frac{4T}{R}\times \frac{4}{3}\pi R^3=\frac{4T}{r_1}\times \frac{4}{3}\pi r_1^3+\frac{4T}{r_2}\times \frac{4}{3}\pi r_2^3$

or $R^2=r_1^2+r_2^2$

$R^2=3^2+4^2$

$R^2=9+16=25$

$R=5cm$