Mechanical Properties of Fluids — Physics STD 11 Science — Question

Consider a bubble of radius R with $\sigma$ the surface tension of liquid. Excess pressure inside the bubble,

P = P_i - P₀

 ($\because$ 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.

$\therefore\text{P}\times\text{R}^2\times\delta\text{R}=8\pi\text{R}(\delta\text{R})\times\sigma$

$\text{P}=\frac{2\sigma}{\text{R}}$

2. 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.,

$\frac{\text{P}}{\rho}+\frac{\text{V}^2}{2}+\text{gh}=\text{constant}$

$\Rightarrow\frac{\text{P}}{\rho\text{g}}+\frac{\text{V}^2}{2\text{q}}+\text{h}=\text{constant}$