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An air bubble of radius 'r' doubles its radius as it rises from a depth 'h' to the surface of the lake at a constant temperature. If the atmospheric pressure is equal to 10 m height of the water column, neglecting surface tension the value of 'h' is:
- a)90 m
- b)70 m
- c)60 m
- d)None of these
Correct answer is option 'B'. Can you explain this answer?
Verified Answer
An air bubble of radius r doubles its radius as it rises from a depth ...
Concept:
Find the expression for both volume and pressure at the bottom and at the surface of the lake. Then at constant temperature try to find the solution with the help of P-V relation.
Hydrostatic Pressure:
Hydrostatic pressure can be defined as the pressure that is exerted by a fluid at equilibrium to an object at a depth due to the gravitational force. It can be given as,
p = ρgh
Calculation:
Given:
Patm = 10 m (height of water column)
At the bottom of the lake, the radius of the air bubble is r.
At the surface of the lake, the radius of the bubble will be 2r.
Volume of the bubble at the bottom of the lake 
Volume of the bubble at the surface of the lake 
Now the atmospheric pressure at the surface is = 2 m.
Atmospheric pressure at the bottom will be
P1 = Patm + Pressure at the bottom of the water at height h
P1 = (10 + h) m
The temperature has been given constant, therefore
P1V1 = P2V2
(10 + h)V = 10 × 8V
10 + h = 80
h = 70 m
Most Upvoted Answer
An air bubble of radius r doubles its radius as it rises from a depth ...
Understanding the Problem
An air bubble rises from a depth h in a lake and doubles its radius as it reaches the surface. We need to find the depth h, given that atmospheric pressure is equivalent to a 10 m water column.
Key Principles
- Boyle's Law: This law states that the pressure and volume of a gas are inversely proportional at a constant temperature (P1V1 = P2V2).
- Pressure Change: As the bubble rises, the pressure decreases from the pressure at depth h (P1) to atmospheric pressure (P2).
Initial Conditions
- Initial Radius: r
- Initial Volume: V1 = (4/3)πr^3
Final Conditions
- Final Radius: 2r
- Final Volume: V2 = (4/3)π(2r)^3 = (32/3)πr^3
Pressure Calculations
1. At Depth h:
- The pressure at depth h can be calculated as:
P1 = ρgh + Patm
Where ρ (density of water) ≈ 1000 kg/m³ and g (acceleration due to gravity) ≈ 10 m/s².
2. At the Surface:
- The pressure at the surface is simply atmospheric pressure:
P2 = Patm
Applying Boyle's Law
Using Boyle's Law:
P1V1 = P2V2
Substituting the volumes:
(ρgh + Patm)(4/3)πr^3 = Patm(32/3)πr^3
The π and (4/3) can be canceled out:
(ρgh + Patm)r^3 = 8Patm
Final Steps
Rearranging gives:
ρgh = 8Patm - Patm
ρgh = 7Patm
Substituting Patm = 1000 kg/m³ × 10 m:
ρgh = 7 × 1000 × 10
Solving for h:
h = (7 × 1000 × 10) / (1000 × 10) = 70 m
Conclusion
Therefore, the value of h is 70 m, confirming option B as the correct answer.
An air bubble rises from a depth h in a lake and doubles its radius as it reaches the surface. We need to find the depth h, given that atmospheric pressure is equivalent to a 10 m water column.
Key Principles
- Boyle's Law: This law states that the pressure and volume of a gas are inversely proportional at a constant temperature (P1V1 = P2V2).
- Pressure Change: As the bubble rises, the pressure decreases from the pressure at depth h (P1) to atmospheric pressure (P2).
Initial Conditions
- Initial Radius: r
- Initial Volume: V1 = (4/3)πr^3
Final Conditions
- Final Radius: 2r
- Final Volume: V2 = (4/3)π(2r)^3 = (32/3)πr^3
Pressure Calculations
1. At Depth h:
- The pressure at depth h can be calculated as:
P1 = ρgh + Patm
Where ρ (density of water) ≈ 1000 kg/m³ and g (acceleration due to gravity) ≈ 10 m/s².
2. At the Surface:
- The pressure at the surface is simply atmospheric pressure:
P2 = Patm
Applying Boyle's Law
Using Boyle's Law:
P1V1 = P2V2
Substituting the volumes:
(ρgh + Patm)(4/3)πr^3 = Patm(32/3)πr^3
The π and (4/3) can be canceled out:
(ρgh + Patm)r^3 = 8Patm
Final Steps
Rearranging gives:
ρgh = 8Patm - Patm
ρgh = 7Patm
Substituting Patm = 1000 kg/m³ × 10 m:
ρgh = 7 × 1000 × 10
Solving for h:
h = (7 × 1000 × 10) / (1000 × 10) = 70 m
Conclusion
Therefore, the value of h is 70 m, confirming option B as the correct answer.
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An air bubble of radius r doubles its radius as it rises from a depth h to the surface of the lake at a constant temperature. If the atmospheric pressure is equal to 10 m height of the water column, neglecting surface tension the value of h is:a)90 mb)70 mc)60 md)None of theseCorrect answer is option 'B'. Can you explain this answer?
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