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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?
Most Upvoted Answer
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).
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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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