Mechanical Properties of Fluids — Physics STD 11 Science — Question

1. Inside a liquid drop: Let r = radius of a spherical liquid drop of centre O.

T = surface tension of the liquid.

Let P_i and P₀ be the values of pressure inside and outside the drop.

$\therefore$ Excess pressure inside the liquid drop = P_i­ - P₀.

Let $\Delta\text{r}$ be the increase in its radius due to excess pressure. It has one free surface outside.

$\therefore$ Increase in surface area of the liquid drop

$=4\pi(\text{r}+\Delta\text{r})^2-4\pi\text{r}^2$

$=4\pi[\text{r}^2+(\Delta\text{r})^2+2\text{r}\Delta\text{r}-\text{r}^2]$

$=8\pi\text{r}\ \Delta\text{r}\dots\text{(i)}$ $(\because\Delta\text{r}$ is small $\therefore\Delta\text{r}^2$ is neglected$)$

$\therefore$ Increase in surface energy of the drop is

W = Surface tension × increase in area

$=\text{T}\times8\pi\text{r }\Delta\text{r}\dots\text{(ii)}$

Also W = Force due to excess of pressure × dispplacement

= Excess pressure × Area of drop × increase in raduis

$=(\text{P}_\text{i}-\text{P}_0)4\pi\text{r}^2\Delta\text{r}\dots\text{(iii)}$

$\therefore$ From equations (ii) and (iii), we get

$(\text{P}_\text{i}-\text{P}_0)\times4\pi\text{r}^2\Delta\text{r}$

$=\text{T}\times\text{8}\pi\text{r }\Delta\text{r}$

$\Rightarrow\text{P}_\text{i}-\text{P}_0=\frac{\text{T}}{\text{r}}$

2. Inside a liquid bubble: A liquid bubble has air both inside and outside it and therefore it has two free surfaces.

Thus increase in its surface area

$=2[4\pi(\text{r}+\Delta\text{r})^2-4\pi\text{r}]$

$=2\times8\pi\text{r}\Delta\text{r}=16\pi\text{r}\Delta\text{r}$

$\therefore\text{W}=\text{T}\times16\pi\text{r }\Delta\text{r}\dots\text{(i)}$

Also $\text{W}=(\text{P}_\text{i}-\text{P}_0)4\pi\text{r}^2\times\Delta\text{r}\dots\text{(ii)}$

From equations (i) and (ii), we get

$(\text{P}_\text{i}-\text{P}_0)\times4\pi\text{r}^2\times\Delta\text{r}=\text{T}\cdot16\pi\text{r}\Delta\text{r}$ or $\text{P}_\text{i}-\text{P}_0=\frac{4\text{T}}{\text{r}}.$

3. Inside an air bubble: Air bubble is formed inside liquid, thus air bubble has one free surface inside it and liquid is outside.

If r = radius of air bubble

$\Delta\text{r}=$ increase in its radius due to excess of pressure

$(\text{P}_\text{i}-\text{P}_0)$ inside it.

T = surface tension of the liquid in which bubble is formed.

$\therefore$ Increase in surface area $=8\pi\text{r}\Delta\text{r}.$

$\therefore\text{W}=\text{T}\times8\pi\text{r}\Delta\text{r}$

Also $\text{W}=(\text{P}_\text{i}-\text{P}_0)\times4\pi\text{r}^2\Delta\text{r}$

$\therefore(\text{P}_\text{i}-\text{P}_0)\times4\pi\text{r}^2\Delta\text{r}=\text{T}\times8\pi\text{r }\Delta\text{r}$ or $\text{P}_\text{i}-\text{P}_0=\frac{2\text{T}}{\text{r}}.$