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#capillary rise

11 public questions tagged with this topic.

Equation of Continuity and Flow Rate

A capillary tube of radius 0.25mm is dipped in ethanol (S\=0.0227N/m, ρ\=806kg/m3, cos⁡θ\=1). What is the capillary rise

h = 2Scos⁡θρga. S = 0.0227N/m, ρ = 806kg/m3, g = 9.8m/s2, a = 0.25×10−3m. h = 2×0.0227×1806×9.8×0.25×10−3 = 0.02298m = 2.3cm. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 2.3 cm. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.

Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.

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Fluid Pressure, Pascal's Law and Manometers

A capillary tube of radius 0.3mm is dipped in water (S\=0.0727N/m, ρ\=1000kg/m3, cos⁡θ\=1). What is the capillary rise?

h = 2Scos⁡θρga. S = 0.0727N/m, ρ = 1000kg/m3, g = 10m/s2, a = 0.3×10−3m. h = 2×0.0727×11000×10×0.3×10−3 = 0.04847m = 4.85cm. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 4.8 cm. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.

Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.

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Surface Tension, Capillarity and Excess Pressure

A capillary tube of radius 0.15mm is dipped in water (S\=0.0727N/m, ρ\=1000kg/m3, cos⁡θ\=1). What is the capillary rise?

h = 2Scos⁡θρga. S = 0.0727N/m, ρ = 1000kg/m3, g = 9.8m/s2, a = 0.15×10−3m. h = 2×0.0727×11000×9.8×0.15×10−3 = 0.0989m = 9.89cm. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 9.9 cm. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.

Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.

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Surface Tension, Capillarity and Excess Pressure

A capillary tube of radius 0.4mm is dipped in water (S\=0.0727N/m, ρ\=1000kg/m3, cos⁡θ\=1). What is the capillary rise?

h = 2Scos⁡θρga. S = 0.0727N/m, ρ = 1000kg/m3, g = 9.8m/s2, a = 0.4×10−3m. h = 2×0.0727×11000×9.8×0.4×10−3 = 0.03714m = 3.71cm. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 3.7 cm. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.

Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.

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Viscosity, Stokes' Law and Reynolds Number

A capillary tube of radius 0.65mm is dipped in water (S\=0.0727N/m, ρ\=1000kg/m3, cos⁡θ\=1). What is the capillary rise?

h = 2Scos⁡θρga. S = 0.0727N/m, ρ = 1000kg/m3, g = 9.8m/s2, a = 0.65×10−3m. h = 2×0.0727×11000×9.8×0.65×10−3 = 0.02285m = 2.285cm. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 2.3 cm. This satisfies dimensional consistency and physical conditions given, so option C is scientifically correct.

Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.

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Bernoulli's Theorem and Applications

A capillary tube of radius 0.05mm is dipped in water (S\=0.0727N/m, ρ\=1000kg/m3, cos⁡θ\=1). What is the capillary rise?

h = 2Scos⁡θρga. S = 0.0727N/m, ρ = 1000kg/m3, g = 9.8m/s2, a = 0.05×10−3m. h = 2×0.0727×11000×9.8×0.05×10−3 = 0.2967m = 29.67cm. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 29.7 cm. This satisfies dimensional consistency and physical conditions given, so option C is scientifically correct.

Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.

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Equation of Continuity and Flow Rate

A capillary tube of radius 0.1mm is dipped in water (S\=0.073N/m, ρ\=1000kg/m3). What is the height of capillary rise? (

Capillary rise: h = 2Scos⁡θρga. S = 0.073N/m, ρ = 1000kg/m3, g = 10m/s2, a = 0.1×10−3m, cos⁡θ = 1. h = 2×0.0731000×10×0.1×10−3 = 0.146m = 14.6cm. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 14.6 cm. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.

Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.

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Equation of Continuity and Flow Rate

A capillary tube of radius 0.35mm is dipped in ethanol (S\=0.0227N/m, ρ\=806kg/m3, cos⁡θ\=1). What is the capillary rise

h = 2Scos⁡θρga. S = 0.0227N/m, ρ = 806kg/m3, g = 10m/s2, a = 0.35×10−3m. h = 2×0.0227×1806×10×0.35×10−3 = 0.0161m = 1.61cm. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 1.6 cm. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.

Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.

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Equation of Continuity and Flow Rate

A capillary tube of radius 0.45mm is dipped in water (S\=0.0727N/m, ρ\=1000kg/m3, cos⁡θ\=1). What is the capillary rise?

h = 2Scos⁡θρga. S = 0.0727N/m, ρ = 1000kg/m3, g = 9.8m/s2, a = 0.45×10−3m. h = 2×0.0727×11000×9.8×0.45×10−3 = 0.03298m = 3.3cm. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 3.3 cm. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.

Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.

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Equation of Continuity and Flow Rate

A capillary tube of radius 0.55mm is dipped in water (S\=0.0727N/m, ρ\=1000kg/m3, cos⁡θ\=1). What is the capillary rise?

h = 2Scos⁡θρga. S = 0.0727N/m, ρ = 1000kg/m3, g = 10m/s2, a = 0.55×10−3m. h = 2×0.0727×11000×10×0.55×10−3 = 0.02645m = 2.645cm. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 2.7 cm. This satisfies dimensional consistency and physical conditions given, so option C is scientifically correct.

Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.

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