Question

The diameter of the droplet is 0.075 mm. What is the intensity of pressure (N/sq.cm) developed in the droplet by the surface tension of 0.000075 N/mm?

• A. $0.4$
• B. $0.6$
• C. $0.8$
• D. $1$

Answer 1

The correct option is A $0.4$



Let the pressure difference between inside and outside be $\Delta P$ .
Let the droplet be cut into two halves

As there is only one interface between liquid and air,
$\pi d\sigma =\frac{\pi d^2}{4}\times \Delta P$

$\Delta P=\frac{4\sigma }{d}$

Where $\sigma$ = Surface tension
d= diameter of droplet
The smaller is the diameter of the bubble layer, the larger is
the pressure difference.
Note: Data Given:
Surface tension, $\sigma$= 0.000075 N/mm
Diameter of the droplet, (d)=0.075 mm
$\Delta P=\frac{4\times 0.000075}{0.075}$
$=4\times {10}^{-3}N/m{m}^2=0.4N/c{m}^2$
Note: Soap Bubble in Air
In a soap bubble, there are two interfaces between soap film and air. One inside the bubble and the other outside.

$Hence,\left(2\pi r\sigma \right)\times 2=\Delta P\times \pi r^2$ $\Rightarrow \Delta P=\frac{4\sigma }{r}$
Liquid Jet : Consider a liquid jet of dia 'd' and length 'l'