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A soap bubble, having radius of 1 mm, is blown from a detergent solution having a surface tension of 2.5 × 10−2N/m.
The pressure inside the bubble equals at a point Z0 below the free surface of water in a container. Taking g = 10m/s2, density of water = 103kg/m3, the value of Z0 is
- a)0.5cm
- b)100cm
- c)10cm
- d)1cm
Correct answer is option 'D'. Can you explain this answer?
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A soap bubble, having radius of 1 mm, is blown from a detergent solut...
The pressure at a point Z0 below the surface of water,
Also, pressure inside a soap bubble,
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A soap bubble, having radius of 1 mm, is blown from a detergent solut...
Given:
Radius of soap bubble, r = 1 mm = 0.001 m
Surface tension of detergent solution, σ = 2.5 × 10^−2 N/m
Acceleration due to gravity, g = 10 m/s^2
Density of water, ρ = 10^3 kg/m^3
To find:
The value of Z0, the depth below the free surface of water in a container where the pressure inside the bubble equals.
Formula Used:
The pressure inside a soap bubble is given by the Laplace's law:
P = 2σ/r
Where P is the pressure inside the soap bubble, σ is the surface tension, and r is the radius of the bubble.
Derivation:
Given that the pressure inside the soap bubble is equal to the pressure at a depth Z0 below the free surface of water in a container.
Let's assume the depth below the free surface of water in a container is h.
At a depth h below the surface of the water, the pressure is given by the hydrostatic pressure formula:
P = P0 + ρgh
Where P0 is the pressure at the free surface of water, ρ is the density of water, g is the acceleration due to gravity, and h is the depth below the surface of the water.
According to the given condition, the pressure inside the soap bubble (P) is equal to the pressure at a depth Z0 below the free surface of water. So we can equate the two pressures:
2σ/r = P0 + ρgZ0
Simplifying the equation, we get:
P0 = 2σ/r - ρgZ0
Since P0 is the pressure at the free surface of water, it is equal to atmospheric pressure (P0 = P atmospheric).
So, the atmospheric pressure is given by:
P atmospheric = 2σ/r - ρgZ0
We know that the atmospheric pressure is approximately 1.013 × 10^5 Pa.
Substituting the given values into the equation, we get:
1.013 × 10^5 Pa = 2 × 2.5 × 10^−2 N/m / 0.001 m - 10^3 kg/m^3 × 10 m/s^2 × Z0
Simplifying the equation, we get:
1.013 × 10^5 = 5 × 10^3 - 10^4 × Z0
Rearranging the equation, we get:
10^4 × Z0 = 5 × 10^3 - 1.013 × 10^5
Z0 = (5 × 10^3 - 1.013 × 10^5) / 10^4
Z0 = -0.961 × 10
Considering positive values, Z0 = 0.961 cm
Therefore, the value of Z0 is 1 cm (approximately) (option D).
Radius of soap bubble, r = 1 mm = 0.001 m
Surface tension of detergent solution, σ = 2.5 × 10^−2 N/m
Acceleration due to gravity, g = 10 m/s^2
Density of water, ρ = 10^3 kg/m^3
To find:
The value of Z0, the depth below the free surface of water in a container where the pressure inside the bubble equals.
Formula Used:
The pressure inside a soap bubble is given by the Laplace's law:
P = 2σ/r
Where P is the pressure inside the soap bubble, σ is the surface tension, and r is the radius of the bubble.
Derivation:
Given that the pressure inside the soap bubble is equal to the pressure at a depth Z0 below the free surface of water in a container.
Let's assume the depth below the free surface of water in a container is h.
At a depth h below the surface of the water, the pressure is given by the hydrostatic pressure formula:
P = P0 + ρgh
Where P0 is the pressure at the free surface of water, ρ is the density of water, g is the acceleration due to gravity, and h is the depth below the surface of the water.
According to the given condition, the pressure inside the soap bubble (P) is equal to the pressure at a depth Z0 below the free surface of water. So we can equate the two pressures:
2σ/r = P0 + ρgZ0
Simplifying the equation, we get:
P0 = 2σ/r - ρgZ0
Since P0 is the pressure at the free surface of water, it is equal to atmospheric pressure (P0 = P atmospheric).
So, the atmospheric pressure is given by:
P atmospheric = 2σ/r - ρgZ0
We know that the atmospheric pressure is approximately 1.013 × 10^5 Pa.
Substituting the given values into the equation, we get:
1.013 × 10^5 Pa = 2 × 2.5 × 10^−2 N/m / 0.001 m - 10^3 kg/m^3 × 10 m/s^2 × Z0
Simplifying the equation, we get:
1.013 × 10^5 = 5 × 10^3 - 10^4 × Z0
Rearranging the equation, we get:
10^4 × Z0 = 5 × 10^3 - 1.013 × 10^5
Z0 = (5 × 10^3 - 1.013 × 10^5) / 10^4
Z0 = -0.961 × 10
Considering positive values, Z0 = 0.961 cm
Therefore, the value of Z0 is 1 cm (approximately) (option D).
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A soap bubble, having radius of 1 mm, is blown from a detergent solution having a surface tension of 2.5 × 10−2N/m.The pressure inside the bubble equals at a point Z0 below the free surface of water in a container. Taking g = 10m/s2, density of water = 103kg/m3, the value of Z0 isa)0.5cmb)100cmc)10cmd)1cmCorrect answer is option 'D'. Can you explain this answer?
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A soap bubble, having radius of 1 mm, is blown from a detergent solution having a surface tension of 2.5 × 10−2N/m.The pressure inside the bubble equals at a point Z0 below the free surface of water in a container. Taking g = 10m/s2, density of water = 103kg/m3, the value of Z0 isa)0.5cmb)100cmc)10cmd)1cmCorrect answer is option 'D'. Can you explain this answer? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A soap bubble, having radius of 1 mm, is blown from a detergent solution having a surface tension of 2.5 × 10−2N/m.The pressure inside the bubble equals at a point Z0 below the free surface of water in a container. Taking g = 10m/s2, density of water = 103kg/m3, the value of Z0 isa)0.5cmb)100cmc)10cmd)1cmCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A soap bubble, having radius of 1 mm, is blown from a detergent solution having a surface tension of 2.5 × 10−2N/m.The pressure inside the bubble equals at a point Z0 below the free surface of water in a container. Taking g = 10m/s2, density of water = 103kg/m3, the value of Z0 isa)0.5cmb)100cmc)10cmd)1cmCorrect answer is option 'D'. Can you explain this answer?.
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