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)rhog) + (2T)/(r_(1)) = P_(1)`
and the total pressure inside the bubble at depth `h_(2)` is `= (P + h_(2)rhog) + (2T)/(r_(2)) = P_(2)`
Now, according to Boyle's Law
`P_(1)V_(1) = P_(2)V_(2)` where `V_(1) = (4)/(3) pir_(1)^(3)`, and `V_(2) = (4)/(3) pir_(2)^(3)`
Hence we get
`[(P + h_(1)rhog) + (2T)/(r_(1))] (4)/(3) pir_(1)^(3) = [(P + h_(1)rhog) + (2T)/(r_21))] (4)/(3) pir_(2)^(3)`
`[(P + h_(1)rhog) + (2T)/(r_(1))]r_(1)^(3) = [(P + h_(1)rhog) + (2T)/(r_(1))]r_(2)^(3)`
Given the : `h_(1) = 100 cm, r_(1) = 0.1 mm = 0.1 mm = 0.01 cm, r_(2) = 0.126 mm = 0.0126 cm, T = 567 "dyne"//"cm", P = 76 cm`
of mercury. Substituting all the values, we get
`h_(2) = 9.48 cm`.