a large bubble rises from the bottom of the lake to the surface and its radius become 4times, find the depth of lake if atmospheric pressure is equal to that of column of water height H
Answers
Explanation:
ANSWER
It is given that the atmospheric pressure is,
P
o
=ρgH, where ρ is the density of water.
Let the depth of the lake be d, the pressure at this depth would be,
P = P
o
+ρgd
Hence, P = ρg(d+H)
Since, the surrounding temperature is constant, we can assume that the process takes place isothermally. Therefore, we can apply the Boyle's Law
P
1
V
1
=P
2
V
2
Here, 1 denotes the water at depth d and 2 denotes the surface of water.
Hence, P
1
=ρg(d+H)
P
2
=ρgH
V
2
=8V
1
(since, radius is doubled, volume becomes 8 times)
Substituting the values, in Boyle's Law,
d = 8H - H = 7H
Answer:
Explanation:
11th
Physics
Mechanical Properties of Fluids
Variation of Pressure With Depth
When a large bubble rises f...
PHYSICS
When a large bubble rises from the bottom of a lake to the surface, its radius doubles. If atmospheric pressure is equal to that of column of water height H, then the depth of lake is :-
December 27, 2019
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Tamana Mouni
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ANSWER
It is given that the atmospheric pressure is,
P
o
=ρgH, where ρ is the density of water.
Let the depth of the lake be d, the pressure at this depth would be,
P = P
o
+ρgd
Hence, P = ρg(d+H)
Since, the surrounding temperature is constant, we can assume that the process takes place isothermally. Therefore, we can apply the Boyle's Law
P
1
V
1
=P
2
V
2
Here, 1 denotes the water at depth d and 2 denotes the surface of water.
Hence, P
1
=ρg(d+H)
P
2
=ρgH
V
2
=8V
1
(since, radius is doubled, volume becomes 8 times)
Substituting the values, in Boyle's Law,
d = 8H - H = 7H