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#Bernoulli equation

3 public questions tagged with this topic.

Viscosity, Stokes' Law and Reynolds Number

Water (ρ\=1000kg/m3) flows horizontally at 3.2m/s with pressure 1.85×105Pa. If the speed increases to 5.8m/s, what is th

Bernoulli’s equation: P1+12ρv12 = P2+12ρv22. P1 = 1.85×105Pa, v1 = 3.2m/s, v2 = 5.8m/s, ρ = 1000kg/m3. P2 = 1.85×105+12×1000(10.24−33.64) = 1.85×105−11700 = 1.733×105Pa. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 1.73 × 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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Bernoulli's Theorem and Applications

Water (ρ\=1000kg/m3) flows horizontally at 4m/s with pressure 1.6×105Pa. If the speed increases to 7m/s, what is the new

Bernoulli’s equation: P1+12ρv12 = P2+12ρv22. P1 = 1.6×105Pa, v1 = 4m/s, v2 = 7m/s, ρ = 1000kg/m3. P2 = 1.6×105+12×1000(16−49) = 1.6×105−16500 = 1.435×105Pa. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 1.435 × 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

Water (ρ\=1000kg/m3) flows horizontally at 2.8m/s with pressure 1.9×105Pa. If the speed increases to 5.2m/s, what is the

Bernoulli’s equation: P1+12ρv12 = P2+12ρv22. P1 = 1.9×105Pa, v1 = 2.8m/s, v2 = 5.2m/s, ρ = 1000kg/m3. P2 = 1.9×105+12×1000(7.84−27.04) = 1.9×105−9600 = 1.804×105Pa. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 1.80 × 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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