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.

Answer

C, D

Explanation

The total pressure inside the bubble at depth `h_(1)` is (p is atmoshpheric 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)rho g)+(2T)/(r_2)=p_(2)`
Now, according to Boyal's law
`p_(1)v_(1)=p_(2)v_(2)`
where, `V_(1)=(4)/(3)pi r_(1)^(3)` and `V_(2)=(4)/(3)pir_(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)pir_(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)=0.1cm,r_(2)=0.126 mm=0.0126cm, T=576 "dyne"//cm, p=76 cm`of mercury. Substituting all the values, we get
`h_(2)=9.48 cm`.