Question
Question: An air bubble which is lying just below the surface of water. The surface tension of water is given ...
An air bubble which is lying just below the surface of water. The surface tension of water is given as 7×10−3Nm−1 and the atmospheric pressure is 1.013×105Nm−2. If the radius of bubble is 1mm, then the pressure inside the bubble will be given as,
A.1.0270×105PaB.1.0160×105PaC.1.0144×105PaD.1.0131×105Pa
Solution
Since it is said that the bubble is just below the surface of water, we can suppose that the pressure outside the bubble will be similar as atmospheric pressure. Therefore the pressure inside the bubble will be greater than that of the outside.
Formula used:
Pin−Pout=R2T
Where T be the surface tension of water and R be the universal gas constant.
Complete answer:
As the bubble is said to be just beneath the surface of the water, we can assume safely that the pressure outside the bubble is similar to the atmospheric pressure. Hence the pressure inside the bubble will be greater than the pressure outside the bubble,
So that we can write it in the form of an equation,
Pin−Pout=R2T
Where T be the surface tension of water and R be the radius of bubbles.
It is given in the question that,
The surface tension of the water is,
T=7×10−3Nm−1
And the radius of the bubble is given as,
R=1mm
And also the outside pressure is,
Pout=1.013×105Nm−2
Substituting this in the equation will give,