13 public questions tagged with this topic.
Using Bernoulli’s equation: P1+12ρv12 = P2+12ρv22 (height constant). P1 = 1.5×105Pa, v1 = 3m/s, v2 = 6m/s, ρ = 1000kg/m3. P2 = P1+12ρ(v12−v22) = 1.5×105+12×1000(9−36). P2 = 1.5×105−13500 = 1.365×105Pa.
Ref: NCBI Bookshelf, Ideal Gas Law: P1V1/T1 = P2V2/T2.
Bernoulli’s principle states that in steady flow, an increase in kinetic energy (12ρv2) corresponds to a decrease in pressure energy, as total energy is conserved along a horizontal streamline. As per NCERT, applying relevant law/formula with correct units and sign convention leads to Because of energy conservation. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.
Ref: NCBI Bookshelf, Ideal Gas Law: P1V1/T1 = P2V2/T2.
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.
Continuity equation: A1v1 = A2v2. A1 = 0.04m2, v1 = 2.5m/s, A2 = 0.01m2. v2 = A1v1A2 = 0.04×2.50.01 = 10m/s. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 10 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.
Using the equation of continuity: A1v1 = A2v2. A1 = 0.02m2, v1 = 2m/s, A2 = 0.01m2. v2 = A1v1A2 = 0.02×20.01 = 4m/s. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 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.
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.
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.
The aerofoil shape causes air to move faster over the top than underneath, reducing pressure above (Bernoulli’s principle) and creating a pressure difference that lifts the plane, known as dynamic lift.
Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.
Bernoulli’s equation: P1+12ρv12 = P2+12ρv22. P1 = 2.0×105Pa, v1 = 4.5m/s, v2 = 7.0m/s, ρ = 1000kg/m3. P2 = 2.0×105+12×1000(20.25−49) = 2.0×105−14375 = 1.85625×105Pa. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 1.86 × 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.
