Air Bubble Equilibrium in Water

What is the depth at which an air bubble of radius 0.0003 m will remain in equilibrium in water?

Given surface tension of water=7* 10-2 N/m, g= 9.8m/s-².

Answer:

The depth at which an air bubble of radius 0.0003 m will remain in equilibrium in water is 0.047 m.

It is given that the radius of the air bubble is 0.0003 m and the surface tension of water is 7* 10-2 N/m. To find the depth at which the air bubble will remain in equilibrium in water, we need to consider the excess pressure inside the air bubble. The excess pressure can be calculated using the formula:

P = 2T / R

Where P is the excess pressure, T is the surface tension, and R is the radius of the air bubble.

Additionally, the pressure due to a height in water is given by:

P = ρgh

Where ρ is the density of water, g is the acceleration due to gravity, and h is the depth.

By equating these two pressures, we can solve for the depth h:

ρgh = 2T / R

After substituting the given values and solving the equation, we find that the depth h is 0.047 m.

Therefore, the air bubble of radius 0.0003 m will remain in equilibrium at a depth of 0.047 m in water.