A minute spherical air bubble is rising slowly through a column of mercury contained in a deep jar. If the radius of the bubble at a depth of `100 cm` is` 0.1mm`,Calculate its depth where its radius is `0.126mm`, given that the surface tension of mercury is `567 dyne//cm`. Assume that the atmosphere pressure is `76cm` of mercury.

Explanation

The total pressure inside the bubble at depth `h_(1)` is ( P is atmospheric pressure )
`= ( P + h_(1) rho g) + ( 2T)/( r_(1)) = P_(1)`
and the total pressure inside the bubble at depth `h_(2)` is `= ( P + h_(2) rho g) + ( 2T)/( r_(2)) = P_(2)`
Now, according to Boyle's Law
`P_(1) V_(1) = P_(2) V_(2)`
where `V_(1) = ( 4)/( 3) pi r_(1)^(3)`, and `V_(2) = ( 4)/( 3) pi r_(2)^(3)`
Hence we get
`[ ( P + h_(1) rho g) + ( 2T)/( r_(1)]( 4)/( 3) pi r_(1)^(3) = [ ( P + h_(2) rho g)+ ( 2T)/( r_(2))]( 4)/( 3) pi r_(2)^(3)`
or, `[ ( P+ h_(1) rho g) + ( 2T)/( r_(1))]r_(1)^(3) = [ ( P+ h_(2)rho g )|+ ( 2T)/( r_(2))]r_(2)^(3)`
Given that `: h_(1) = 100cm, r_(1) = 0.1m m = 0.01 cm, r_(2) = 0.126 m m = 0.0126 cm, T = 567 ` dyne `//` cm, P = 76 cm of mercury . Substituting all the values, we get
`h_(2) = 9.48cm`