Skip to content
Login Register

#fluid flow

19 public questions tagged with this topic.

Fluid Pressure, Pascal's Law and Manometers

Water flows at 2.9m/s at 3.2m height with pressure 1.9×105Pa. What is the pressure at 1.8m height with speed 4.0m/s? (ρ\

Bernoulli’s: P1+12ρv12+ρgh1 = P2+12ρv22+ρgh2. Left: 1.9×105+12×1000×8.41+1000×10×3.2 = 1.9×105+4205+32000 = 2.26205×105. Right: P2+12×1000×16+1000×10×1.8 = P2+8000+18000 = P2+26000. 2.26205×105 = P2+26000, P2 = 2.00205×105Pa. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 2.00 × 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.

View Details
Surface Tension, Capillarity and Excess Pressure

A spray tube (7cm2) has 35 holes of diameter 0.8mm. If the flow speed is 3.0m/min, what is the ejection speed?

A1v1 = A2v2, v1 = 3.0m/min = 0.05m/s, A1 = 7×10−4m2. Hole area: Ah = π(0.4×10−3)2, total A2 = 35×π×1.6×10−7≈1.759×10−5m2. v2 = 7×10−4×0.051.759×10−5≈1.99m/s. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 2.0 m/s. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.

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

View Details
Viscosity, Stokes' Law and Reynolds Number

A spray tube (5cm2) has 25 holes of diameter 1.3mm. If the flow speed is 2.7m/min, what is the ejection speed?

A1v1 = A2v2, v1 = 2.7m/min = 0.045m/s, A1 = 5×10−4m2. Hole area: Ah = π(0.65×10−3)2, total A2 = 25×π×4.225×10−7≈3.318×10−5m2. v2 = 5×10−4×0.0453.318×10−5≈0.678m/s. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 0.7 m/s. This satisfies dimensional consistency and physical conditions given, so option C is scientifically correct.

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

View Details
Bernoulli's Theorem and Applications

A spray tube (9cm2) has 45 holes of diameter 0.6mm. If the flow speed is 2.0m/min, what is the ejection speed?

A1v1 = A2v2, v1 = 2.0m/min = 0.0333m/s, A1 = 9×10−4m2. Hole area: Ah = π(0.3×10−3)2, total A2 = 45×π×9×10−8≈1.272×10−5m2. v2 = 9×10−4×0.03331.272×10−5≈2.36m/s. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 2.4 m/s. This satisfies dimensional consistency and physical conditions given, so option C is scientifically correct.

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

View Details
Bernoulli's Theorem and Applications

Water flows at 1.5m/s at 2.5m height with pressure 1.8×105Pa. What is the pressure at 0.5m height with speed 2.5m/s? (ρ\

Bernoulli’s: P1+12ρv12+ρgh1 = P2+12ρv22+ρgh2. Left: 1.8×105+12×1000×2.25+1000×10×2.5 = 1.8×105+1125+25000 = 2.06125×105. Right: P2+12×1000×6.25+1000×10×0.5 = P2+3125+5000 = P2+8125. 2.06125×105 = P2+8125, P2 = 1.98×105Pa. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 1.98 × 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.

View Details
Bernoulli's Theorem and Applications

Water (ρ\=1000kg/m3) flows horizontally at 5.0m/s with pressure 2.1×105Pa. If the speed increases to 7.5m/s, what is the

Bernoulli’s equation: P1+12ρv12 = P2+12ρv22. P1 = 2.1×105Pa, v1 = 5.0m/s, v2 = 7.5m/s, ρ = 1000kg/m3. P2 = 2.1×105+12×1000(25−56.25) = 2.1×105−15625 = 1.94375×105Pa. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 1.94 × 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.

View Details
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.

View Details
Fluid Pressure, Pascal's Law and Manometers

What assumption does Bernoulli’s principle make about fluid flow that limits its application to real fluids?

Bernoulli’s principle assumes zero viscosity (non-viscous flow), ignoring frictional losses that occur in real fluids, where viscosity converts kinetic energy into heat, reducing its applicability. As per NCERT, applying relevant law/formula with correct units and sign convention leads to Zero viscosity. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.

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

View Details
Fluid Pressure, Pascal's Law and Manometers

Water flows at 3.5m/s at 4.5m height with pressure 1.5×105Pa. What is the pressure at 2.5m height with speed 4.5m/s? (ρ\

Bernoulli’s: P1+12ρv12+ρgh1 = P2+12ρv22+ρgh2. Left: 1.5×105+12×1000×12.25+1000×9.8×4.5 = 1.5×105+6125+44100 = 2.00225×105. Right: P2+12×1000×20.25+1000×9.8×2.5 = P2+10125+24500 = P2+34625. 2.00225×105 = P2+34625, P2 = 1.656×105Pa. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 1.65 × 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.

View Details