Answer:
h = 0.047 m
Explanation:
It is given that,
Radius of an air bubble is 0.0003 m
The surface tension of water is [tex]7\times 10^{-2}\ N/m[/tex]
We need to find the the depth at which an air bubble of radius 0.0003 m will remain in equilibrium in water. Let it is given by h.
The excess pressure inside air bubble Is given by the formula as follows :
[tex]P=\dfrac{2T}{R}[/tex] .....(1)
T is surface tension
Also, pressure due to a height is given by :
[tex]P=\rho gh[/tex]
So, equation (1) becomes :
[tex]\rho g h=\dfrac{2T}{R}[/tex]
So,
[tex]h=\dfrac{2T}{\rho gR}[/tex]
[tex]h=\dfrac{2\times 7\times 10^{-2}}{10^3\times 9.8\times 0.0003}\\\\h=0.047\ m[/tex]
So, the depth is 0.047 m.
