Question
- Derive an expression for the excess pressure inside a soap bubble.
- State Bernoulli's theorem.
✓
Answer
Consider a bubble of radius R with $\sigma$ the surface tension of liquid. Excess pressure inside the bubble,
$P = P_i - P_0$
($\because$ air bubble has only one free surface)
$\delta\text{R}=$ Small increase in radius of bubble due to excess pressure
Work done,
W = Force × Displacement
= (Excess pressure × Area) × Increase in radius
$=\text{P}\times4\pi\text{R}^2\times\delta\text{R}$
Increase in surface area of bubble,
= Final surface area - Initial surface area
$=4\pi(\text{R}+\delta\text{R})^2-4\pi\text{R}^2$
$=8\pi\text{R}(\delta\text{R})$ $($Neglecting $\delta\text{R}^2)$
$\therefore\text{P}\times4\pi\text{R}^2\times\delta\text{R}=8\pi\text{R}(\delta\text{R})\times\sigma$
Increase in P.E. = Increase in surface area × Surface tension
$=8\pi\text{R}(\delta\text{R})\times\sigma$
Since the drop is in equilibrium.
$\therefore\text{P}\times\text{R}^2\times\delta\text{R}=8\pi\text{R}(\delta\text{R})\times\sigma$
$\text{P}=\frac{2\sigma}{\text{R}}$
- Bernoulli's Theorem: For an incompressible, non-viscous, irrotational liquid having streamlined flow, the sum of the pressure energy, kinetic energy and potential energy per unit mass is a constant, i.e.,
$\Rightarrow\frac{\text{P}}{\rho\text{g}}+\frac{\text{V}^2}{2\text{q}}+\text{h}=\text{constant}$
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