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#mercury drop

3 public questions tagged with this topic.

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.

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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.

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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