Question

Two soap bubble of radius 3 cm and 4 cm combine
together in isothermal condition. The radius of
resultant soap bubble is
(1) 5 cm
(2) 7 cm
(3) 1 cm
(4) V91 cm​

Answer 1

Answer:

Option (1) 5 cm

Explanation:

if the radius of the soap bubble is r, T is surface tension then

Pressure inside the soap bubble = 4T/r

Let the combined radius of the bubble is R

If two bubbles of radius r₁ = 3 cm and r₂ = 4 cm combine to form a bubble of radius R then

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

$\implies R^2=r_1^2+r_2^2$

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

$\implies R^2=25$

$\implies R=5\text{ cm}$

The radius of the resultant soap bubble is 5 cm

Hope this answer is helpful.

Answer 2

Answer:

5 cm

Option 1

Explanation:

For isothermal combination of bubbles, the resultant bubble has the radius:

$r(new) = \sqrt{r1 + r2}$

where r1 and r2 are the radii of the combining bubbles.

Here,

r1 = 3 cm

r2 = 4 cm

So, the new radius is:

$r(new) = \sqrt{ {3 }^{2} + {4}^{2} } = \sqrt{ {5 }^{2} } = 5cm$

Hope it helps.