If the excess pressure inside a soap bubble of radius $r_{1}$ in air is equal to the excess pressure inside air bubble of radius $r_{2}$ inside the soap solution, then $r_{1} : r_{2}$ is

2 : 1

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

1 : 2

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

1 : 4

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

\sqrt{2} : 1

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

1 : \sqrt{2}

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

Open in App

Solution

The correct option is B $2 : 1$
When the soap bubble of radius $r_{1}$ in air, the excess pressure
$Δ p_{1} = \frac{4 T}{r_{1}}$
When the air bubble of radius $r_{2}$ inside the soap solution, the excess pressure
$Δ p_{2} = \frac{2 T}{r_{2}}$
$∴ \frac{Δ p_{1}}{Δ p_{2}} = \frac{4 T}{r_{1}} / \frac{2 T}{r_{2}}$
$Δ p_{1}$ and $Δ p_{2}$ are same
$\frac{Δ p_{1}}{Δ p_{2}} = 1$
$∴ 1 = \frac{2 r_{2}}{r_{1}}$
or $\frac{r_{1}}{r_{2}} = 2$
or $r_{1} : r_{2} = 2 : 1$

Suggest Corrections

Similar questions

Find the excess pressure inside (a) a drop of mercury of radius 2 mm (b) a soap bubble of radius 4 mm and (c) an air bubble of radius 4 mm formed inside a tank of water. Surface tension of mercury , soap solution and water are $0.46 N m^{- 1}, 0.03 N m^{- 1}$ and $0.076 N m^{- 1}$ respectively.

Q. A soap bubble of radius '

r

' is formed in air. The excess pressure inside the bubble is:

What is the excess pressure inside a bubble of soap solution of radius 5.00 mm, given that the surface tension of soap solution at the temperature $(20^{\circ} C)$ is $2.50 \times 10^{- 2} N m^{- 1}$ ? If an air bubble of the same dimension were formed at depth of 40.0 cm inside a container containing the soap solution (of relative density 1.20), what would be the pressure inside the bubble? (1 atmospheric pressure is $1.01 \times 10^{5} P a)$ .