#excess pressure

Surface Tension, Capillarity and Excess Pressure

What is the excess pressure inside a mercury drop of radius 4mm at 20∘C? (Surface tension = 0.4355N/m)

Excess pressure: ΔP = 2Sr. S = 0.4355N/m, r = 4×10−3m. ΔP = 2×0.43554×10−3 = 217.75Pa. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 218 Pa. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.

Ref: NCBI Bookshelf, Ideal Gas Law: P1V1/T1 = P2V2/T2.

View Details

Bernoulli's Theorem and Applications

A soap bubble of radius 8mm has a surface tension of 0.027N/m. What is the excess pressure inside?

Excess pressure in a bubble: ΔP = 4Sr. S = 0.027N/m, r = 8×10−3m. ΔP = 4×0.0278×10−3 = 13.5Pa. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 13.5 Pa. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.

Ref: NCBI Bookshelf, Ideal Gas Law: P1V1/T1 = P2V2/T2.

View Details

Bernoulli's Theorem and Applications

What is the excess pressure inside a mercury drop of radius 1.5mm at 20∘C? (Surface tension = 0.4355N/m)

Excess pressure: ΔP = 2Sr. S = 0.4355N/m, r = 1.5×10−3m. ΔP = 2×0.43551.5×10−3 = 580.67Pa. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 580 Pa. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.

Ref: NCBI Bookshelf, Ideal Gas Law: P1V1/T1 = P2V2/T2.

View Details

Equation of Continuity and Flow Rate

A soap bubble of radius 7mm has a surface tension of 0.028N/m. What is the excess pressure inside?

Excess pressure in a bubble: ΔP = 4Sr. S = 0.028N/m, r = 7×10−3m. ΔP = 4×0.0287×10−3 = 16Pa. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 16 Pa. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.

Ref: NCBI Bookshelf, Ideal Gas Law: P1V1/T1 = P2V2/T2.

View Details

Equation of Continuity and Flow Rate

What is the excess pressure inside a mercury drop of radius 2mm at 20∘C? (Surface tension of mercury = 0.465N/m)

Excess pressure in a drop: ΔP = 2Sr. S = 0.465N/m, r = 2×10−3m. ΔP = 2×0.4652×10−3 = 465Pa. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 465 Pa. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.

Ref: NCBI Bookshelf, Ideal Gas Law: P1V1/T1 = P2V2/T2.

View Details

Fluid Pressure, Pascal's Law and Manometers

A soap bubble of radius 11mm has a surface tension of 0.027N/m. What is the excess pressure inside?

Excess pressure in a bubble: ΔP = 4Sr. S = 0.027N/m, r = 11×10−3m. ΔP = 4×0.02711×10−3 = 9.82Pa. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 10 Pa. This satisfies dimensional consistency and physical conditions given, so option C is scientifically correct.

Ref: NCBI Bookshelf, Ideal Gas Law: P1V1/T1 = P2V2/T2.

View Details

Viscosity, Stokes' Law and Reynolds Number

What is the excess pressure inside a water drop of radius 1.8mm at 20∘C? (Surface tension = 0.0727N/m)

Excess pressure: ΔP = 2Sr. S = 0.0727N/m, r = 1.8×10−3m. ΔP = 2×0.07271.8×10−3 = 80.78Pa. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 80 Pa. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.

Ref: NCBI Bookshelf, Ideal Gas Law: P1V1/T1 = P2V2/T2.

View Details

Viscosity, Stokes' Law and Reynolds Number

What is the excess pressure inside an ethanol drop of radius 2.2mm at 20∘C? (Surface tension = 0.0227N/m)

Excess pressure: ΔP = 2Sr. S = 0.0227N/m, r = 2.2×10−3m. ΔP = 2×0.02272.2×10−3 = 20.64Pa. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 22 Pa. This satisfies dimensional consistency and physical conditions given, so option C is scientifically correct.

Ref: NCBI Bookshelf, Ideal Gas Law: P1V1/T1 = P2V2/T2.

View Details

Bernoulli's Theorem and Applications

What is the excess pressure inside a helium drop of radius 1mm at −270∘C? (Surface tension = 0.000239N/m)

Excess pressure: ΔP = 2Sr. S = 0.000239N/m, r = 1×10−3m. ΔP = 2×0.0002391×10−3 = 0.478Pa. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 0.48 Pa. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.

Ref: NCBI Bookshelf, Ideal Gas Law: P1V1/T1 = P2V2/T2.

View Details

Bernoulli's Theorem and Applications

What is the excess pressure inside a water drop of radius 2.5mm at 20∘C? (Surface tension = 0.0727N/m)

Excess pressure: ΔP = 2Sr. S = 0.0727N/m, r = 2.5×10−3m. ΔP = 2×0.07272.5×10−3 = 58.16Pa. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 58 Pa. This satisfies dimensional consistency and physical conditions given, so option C is scientifically correct.

Ref: NCBI Bookshelf, Ideal Gas Law: P1V1/T1 = P2V2/T2.

View Details

Equation of Continuity and Flow Rate

A soap bubble of radius 10mm has a surface tension of 0.025N/m. What is the excess pressure inside?

Excess pressure in a bubble: ΔP = 4Sr. S = 0.025N/m, r = 10×10−3m. ΔP = 4×0.02510×10−3 = 10Pa. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 10 Pa. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.

Ref: NCBI Bookshelf, Ideal Gas Law: P1V1/T1 = P2V2/T2.

View Details

Equation of Continuity and Flow Rate

A soap bubble of radius 8mm has a surface tension of 0.026N/m. What is the excess pressure inside?

Excess pressure in a bubble: ΔP = 4Sr. S = 0.026N/m, r = 8×10−3m. ΔP = 4×0.0268×10−3 = 13Pa. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 13 Pa. This satisfies dimensional consistency and physical conditions given, so option C is scientifically correct.

Ref: NCBI Bookshelf, Ideal Gas Law: P1V1/T1 = P2V2/T2.

View Details