Excess Pressure in Bubbles
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A soap bubble is given a negative charge, then its radius
- 6 cm
- 5 cm
- 4 cm
- 3 cm
[Take g=10 m/s2]
- 5 cm
- 8 cm
- 1 cm
- 3 cm
What is the excess pressure inside a bubble of soap solution of radius 5.00 mm, given that the surface tension of soap solution at the temperature (20∘C) is 2.50×10–2Nm–1? If an air bubble of the same dimension were formed at depth of 40.0 cm inside a container containing the soap solution (of relative density 1.20), what would be the pressure inside the bubble? (1 atmospheric pressure is 1.01×105Pa).
- decreases
- increases
- remains same
- becomes zero
On heating water, bubbles being formed at the bottom of the vessel detach and rise. Take the bubbles to be spheres of radius R and making a circular contact of radius with the bottom of the vessel. If r << R and the surface tension of water is T, the value of r just before bubbles detach is (density of water is ρw).
R2√ρwgT
R2√2ρwg3T
R2√3ρwgT
R2√ρwg6T
- 150 dyne/cm2
- 3×10−3 dyne/cm2
- 12 dyne/cm2
- 300 dyne/cm2
- 1:9
- 1:3
- 3:1
- 1:27
- 35 N/m2
- 70 N/m2
- 140 N/m2
- 150 N/m2
- 1:16
- 1:64
- 1:4
- 1:8
- 1:27
- 3:1
- 1:3
- 1:9
- 3
- 0.3
- 0.6
- 6
[Take g=10 m/s2]
- 7 mm
- 0.07 mm
- 0.7 mm
- 0.007 mm
- 150 dyne/cm2
- 300 dyne/cm2
- 3×10−3 dyne/cm2
- 12 dyne/cm2
What is the pressure inside the drop of mercury of radius 3.00 mm at room temperature? Surface tension of mercury at that temperature (20∘C) is 4.65×10–1Nm–1 . The atmospheric pressure is 1.01×105 Pa. Also give the excess pressure inside the drop.
- 1.33 cm
- 0.75 cm
- 7.5 cm
- 13.3 cm
Water is flowing in streamline motion through a tube with its axis horizontal. Consider two points A and B in the tube at the same horizontal level.
The pressures at A and B are equal for any shape of the tube.
The pressures are never equal.
The pressures are equal if the tube has a uniform cross section.
The pressures may be equal even if the tube has a non-uniform cross section.
Given: Surface tension of water is 70×10−3 N/m.
- 1 mm
- 2 mm
- 0.1 mm
- 0.2 mm
- 1.47 mm
- 0.73 mm
- 1.45 mm
- 1.46 mm
- 4.8×106 Nm−2
- 5.2×106 Nm−2
- 6.2×106 Nm−2
- 3.1×106 Nm−2
(ρ=0.8 g/cm3) column of height 2 mm, then the surface tension of soap solution will be
[Take g=10 m/s2]
- 0.02 N/m
- 0.04 N/m
- 0.09 N/m
- 0.08 N/m
(Take surface tension of water T=7.2×10−2 N/m and g=10 m/s2)
- 1.11×105 Pa
- 2.11×105 Pa
- 3.5×105 Pa
- 3×105 Pa
- 1 mm
- 0.5 mm
- 0.4 mm
- 0.2 mm
- 1.052×105 Pa
- 1.039×105 Pa
- 1.152×105 Pa
- 1.022×105 Pa
Find the increase in pressure required to decrease the volume of a water sample by 0.01%. Bulk modulus of water = 2.1×109Nm−2
- sin−1(34)
- sin−1(35)
- cos−1(34)
- cos−1(35)
- 4:1
- 1:4
- 2:1
- 1:2