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An air bubble of diameter, d = 4µm is located in water at a depth h = 5.0m Considering standard atmospheric prssure as 1 atm, find the pressure in the air - bubble? (S=0.07 N/m)
Asked by abhisheknand5151 | 10 Jan, 2024, 02:13: PM
Expert Answer
For air-bubble inside water we have only one surface . Hence if Pi is pressure inside bubble and Po be the pressure at the boundary , then we have
where S is surface tension of water and r is radius of bubble .
Pressure at boundary of bubble is sum of atmospheric pressure and pressure due to water column above bubble.
Po = Patm + ( ρ g h )
where Patm is atmospheric pressure , ρ is density of water , g is acceleration due to gravity and h is height of water column above bubble.
Atmospheric pressure Patm = 1.01 × 105 N/m2 .
Po = 1.5 × 105 N/m2
Pressure inside bubble , Pi = Po + [ (2S)/r ]
Pi = [ 1.5 × 105 + ( 2 × 0.07 ) / (2 × 10-6) ] = 2.2 × 105 N/m2
Answered by Thiyagarajan K | 10 Jan, 2024, 04:11: PM
JEE main - Physics
Asked by abhisheknand5151 | 10 Jan, 2024, 02:13: PM
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