Consider an air bubble of radius 2 mm in a liquid at a depth of 5 cm below the free surface. The density of the liquid is `1000 kg//m ^(3)` and the surface tension is `0.1N//m.` Then, find the pressure inside the air bubble is greater than the pressure at the free surface of the liquid. (Take ,` h = 10 m //s^(2))`

Options

A. 500 Pa
B. 600 Pa
C. 700 Pa
D. 800 Pa

Answer

B

Explanation

Gauge pressure due to liquid at a depth of 5 cm
`p = rho gh`
Density, `rho = 1000kg//m^(3), h = 5 cm =0.05cm`
`g = 10 m//s ^(2)`
` p = 100 xx 10 xx 0.05`
` p = 500 Pa`
`implies `Excess pressure inside the air bubble w.r.t outside due to surface tension is `p '= (2S)/(R)`
Surface tension, `S = 0.1 N//m`
Radius, `R = 2mm = 0.002m`
`p'( 2 xx0.1)/(0.002) =100 Pa`
Now, total excess pressure inside the bobble w.r.t surface of liquid, i.e.
`implies p _("total ") = p+ p'`
`=500 + 100 =600 Pa`