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Standard VIII

Chemistry

Surface Tension

A soap bubble...

Question

A soap bubble (surface tension T) is charged to a charge density $σ$ . If the pressure inside the bubble is equal to outside pressure in equilibrium, the radius of the bubble is given by

$\frac{σ^{2}}{8 ϵ_{0} T}$

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$\frac{8 ϵ_{0} T}{σ^{2}}$

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$\frac{σ}{\sqrt{8 ϵ_{0} T}}$

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$\frac{\sqrt{8 ϵ_{0} T}}{σ}$

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Solution

The correct option is B
$\frac{8 ϵ_{0} T}{σ^{2}}$

For a soap bubble to be in equilibrium there has to be a pressure difference of $\frac{4 T}{r}$ , so

$\frac{4 T}{r} \times Area = \frac{σ^{2}}{2 ϵ_{0}} \times Area$

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Similar questions

Q. A soap bubble of radius 3 cm is charged with surface charge density

σ

such that

\frac{σ^{2}}{ϵ_{0}} = 0.4 S I

units. The excess pressure inside the bubble in SI unit (surface tension of soap solution

= 3 \times 10^{- 3}

N/m) is:

Q. In air, a charged soap bubble of radius 'r' is in equilibrium having outside and inside pressures being equal. The charge on the bubble is ______.

(

ϵ =

permittivity of free space, T=surface tension of soap solution).

Q. In air, a charged soap bubble of radius

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is in equilibrium having outside and inside pressures being equal. The charge on the drop is (

ε_{0} =

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A soap bubble of radius $1.0$ cm is electrically charged so that the pressure inside the bubble is identical with the outside pressure. If the surface tension of the soap solution is $0.028$ N /m, the electric charge on the bubble will be:

Q. The excess pressure across a soap bubble of radius

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p = \frac{4 σ}{r},

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is the surface tension of soap solution. What is the excess pressure across an air bubble of the same radius

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A soap bubble (surface tension T) is charged to a charge density σ. If the pressure inside the bubble is equal to outside pressure in equilibrium, the radius of the bubble is given by

The correct option is B 8ϵ0Tσ2 For a soap bubble to be in equilibrium there has to be a pressure difference of 4Tr, so 4Tr×Area=σ22ϵ0×Area

A soap bubble (surface tension T) is charged to a charge density $σ$ . If the pressure inside the bubble is equal to outside pressure in equilibrium, the radius of the bubble is given by

The correct option is B
$\frac{8 ϵ_{0} T}{σ^{2}}$

For a soap bubble to be in equilibrium there has to be a pressure difference of $\frac{4 T}{r}$ , so

$\frac{4 T}{r} \times Area = \frac{σ^{2}}{2 ϵ_{0}} \times Area$