An air bubble of radius 0 .1, cm is in a liqu | vedclass

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Question

Given: Radius of air bubble,

$r = 0.1\,cm = {10^{ - 3}}m$

Surface tension of liquid,

$S = 0.06\,N/m = 6 	imes {10^{ - 2}}\,N/m$

Dens

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$0.1$

Vedclass · Answer

Given: Radius of air bubble,

$r = 0.1\,cm = {10^{ - 3}}m$

Surface tension of liquid,

$S = 0.06\,N/m = 6 	imes {10^{ - 2}}\,N/m$

Density of liquid, $ho  = {10^3}\,kg/{m^3}$

Excess pressure inside the bubble,

${ho _{exe}} = 1100N{m^{ - 2}}$

Depth of bubble below the liquid surface,

$h=?$

As we know,

${ho _{Excess}} = hho g + \frac{{2s}}{r}$

$ \Rightarrow 1100 = h 	imes {10^3} 	imes 9.8 + \frac{{2 	imes 6 	imes {{10}^{ - 2}}}}{{{{10}^{ - 3}}}}$

$ \Rightarrow 1100 - 9800h + 120$

$ \Rightarrow 9800h = 1100 - 120$

$ \Rightarrow h = \frac{{980}}{{9800}} = 0.1\,m$

An air bubble of radius $0 .1\, cm$ is in a liquid having surface tension $0.06\, N/m$ and density $10^3\, kg/m^3$. The pressure inside the bubble is $1100\, Nm^{-2}$ greater than the atmospheric pressure. At ....... $m$ depth is the bubble below the surface of the liquid ? $(g\, = 9.8\, ms^{- 2})$

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