#total pressure

Equation of Continuity and Flow Rate

What is the total pressure at a depth of 1.8m in oxygen liquid (ρ\=1.43kg/m3) with atmospheric pressure 1.01×105Pa? (Tak

Total pressure: P = Pa+ρgh. Pa = 1.01×105Pa, ρ = 1.43kg/m3, g = 9.8m/s2, h = 1.8m. P = 1.01×105+1.43×9.8×1.8 = 1.01×105+25.21 = 1.0102521×105Pa. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 1.01 × 10⁵ Pa. This satisfies dimensional consistency and physical conditions given, so option A is scientifically correct.

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

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Fluid Pressure, Pascal's Law and Manometers

What is the total pressure at a depth of 3.6m in mercury (ρ\=13.6×103kg/m3) with atmospheric pressure 1.01×105Pa? (Take

Total pressure: P = Pa+ρgh. Pa = 1.01×105Pa, ρ = 13.6×103kg/m3, g = 9.8m/s2, h = 3.6m. P = 1.01×105+13.6×103×9.8×3.6 = 1.01×105+479808 = 5.80808×105Pa. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 5.8 × 10⁵ 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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Viscosity, Stokes' Law and Reynolds Number

What is the total pressure at a depth of 1.6m in air (ρ\=1.29kg/m3) with atmospheric pressure 1.01×105Pa? (Take g\=9.8m/

Total pressure: P = Pa+ρgh. Pa = 1.01×105Pa, ρ = 1.29kg/m3, g = 9.8m/s2, h = 1.6m. P = 1.01×105+1.29×9.8×1.6 = 1.01×105+20.23 = 1.0102023×105Pa. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 1.01 × 10⁵ Pa. This satisfies dimensional consistency and physical conditions given, so option A 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 total pressure at a depth of 2.2m in hydrogen liquid (ρ\=9.0×10−2kg/m3) with atmospheric pressure 1.01×105Pa

Total pressure: P = Pa+ρgh. Pa = 1.01×105Pa, ρ = 9.0×10−2kg/m3, g = 9.8m/s2, h = 2.2m. P = 1.01×105+9.0×10−2×9.8×2.2 = 1.01×105+1.94 = 1.0100194×105Pa. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 1.01 × 10⁵ Pa. This satisfies dimensional consistency and physical conditions given, so option A is scientifically correct.

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

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Equation of Continuity and Flow Rate

A bubble of radius 4.0mm is blown at 60cm depth in a soap solution (ρ\=1.2×103kg/m3, S\=0.025N/m). What is the total pre

Pi = Pa+ρgh+2Sr. Pa = 1.01×105Pa, ρgh = 1.2×103×10×0.6 = 7200Pa. 2Sr = 2×0.0254.0×10−3 = 12.5Pa. Pi = 1.01×105+7200+12.5 = 1.082125×105Pa. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 1.08 × 10⁵ 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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Fluid Pressure, Pascal's Law and Manometers

What is the total pressure at a depth of 2.8m in whole blood (ρ\=1.06×103kg/m3) with atmospheric pressure 1.01×105Pa? (T

Total pressure: P = Pa+ρgh. Pa = 1.01×105Pa, ρ = 1.06×103kg/m3, g = 9.8m/s2, h = 2.8m. P = 1.01×105+1.06×103×9.8×2.8 = 1.01×105+29094.4 = 1.300944×105Pa. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 1.30 × 10⁵ 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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