Question

In Jaeger's method, at the time of bursting of the bubble
A. the internal pressure of the bubble is always greater than the external pressure
B. the internal pressure of the bubble is always equal to the external pressure
C. the internal pressure of the bubble is always less than the external pressure
D. the internal pressure of the bubble is always slightly greater

Answer 1

Hint: The surface tension leads a liquid to acquire minimum surface area. Jaeger’s method is an experiment to calculate the surface tension using pressure. It is based on the principle that the surface tension depends on the pressure difference and the radius of capillary used in the experiment.

Complete step by step solution:
Jaeger's method is used to measure the surface tension. It is based on the principle that the excess pressure inside an air bubble is equal to twice of the surface tension of the liquid in which the bubble forms over the radius of the air bubble.

In this experiment, a capillary of a given radius is dipped into a liquid of which surface tension is to be calculated. A bubble is blown into the capillary.Then
$\sigma =\dfrac{\Delta P\times {{R}_{cap}}}{2}$
Where $\sigma $ is the surface tension of the liquid, $\Delta P$ is the maximum pressure drop and ${{R}_{cap}}$ is the radius of capillary.

So we say that the surface tension depends upon the radius of capillary. So as the size increases , the surface tension inside the bubble . Thus , at the time of bursting the bubble, the pressure inside the bubble increases and the surface tension increases than the pressure outside the bubble.

Thus, option A is the correct answer.

Note: We can find the surface tension by many methods like spinning drop method, pendant drop method, drop volume method, sessile drop method and many more. Jaeger’s method is a measurement technique for determining the surface tension at short surface ages.