InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
As an air bubble rises from the bottom of a large water storage tank to free surface of water, the radius of the air bubble increases from 6 mn to 10 mm. The temperature of the water at the surface is `42^(@)C` and its bottom is `27^(@)C.` Find the depth of the water tank. (Take density of water `=1g cm^(-3), g=10 ms^(-2)`, 1 atmospheric pressure = 760 mm of Hg, density of mercury `=13.6 g cm^(3)` |
|
Answer» (i) Consider the pressure `(P_(1))," volume "(V_(1))" and temperature "(T_(1))` of the air in the bubble in terms of S.I. units. `P_(1)=1 atm = 10.336` m of water `V_(1)=(4)/(3)xxpi r_(1)^(3)=(4)/(3)xxpi((10)/(1000)m)^(3)` `T_(1)=(42+273)K` Find the volume `(V_(2))" and temperature "(T_(2))` of the air bubble at the bottom of the tank. `V_(2)=(4)/(3)xxpi r_(2)^(3)=(4)/(3)xxpi((6)/(1000)m)^(3)` `T_(2)=(27+273)K` Apply the value of `P_(2)` (in terms of pressure exerted by water columns) from (1) Then, the height of the water colomn, `h=(P_(2)-P_(1))` m (ii) `35.273` m |
|