Question

Find the depth of water at which an air of radius 0.003m bubble may remain in equilibrium (surface tension=0.072Nm,gravity=9.8gs²)​

Answer 1

h = 0.047 m

Step-by-step explanation:

It is given that,

Radius of an air bubble is 0.0003 m

The surface tension of water is
`7* 10^(-2)\ N/m`

We need to find the depth at which an air bubble of radius 0.0003 m will remain in equilibrium in water. Let it is given by h.

Pressure due to surface tension is given by :

`P=(2T)/(R)` .....(1)

T is surface tension

Also, pressure due to a height is given by :

`P=\rho gh`

So, equation (1) becomes :

`\rho g h=(2T)/(R)`

So,

`h=(2T)/(\rho gR)`

`h=(2* 7* 10^(-2))/(10^3* 9.8* 0.0003)\\\\h=0.047\ m`

So, the depth is 0.047 m.