Questions covering Bernoulli's theorem, fluid pressure variations, and practical applications such as venturi meters and airfoil lift. Tests understanding of energy conservation in fluid flow.
Pressure in a fluid at rest acts equally in all directions (Pascal’s law) and increases with depth (P = Pa+ρgh), not influenced by the container’s shape, ruling out options involving directionality or shape dependence.
Ref: NCBI Bookshelf, Ideal Gas Law: P1V1/T1 = P2V2/T2.
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.
Hydrostatic pressure (P = ρgh) depends on depth, not the container’s shape or fluid velocity (which applies to dynamic flow). The incorrect statement is that it depends on container shape.
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.
ΔP⋅A = mg. mg = 4.0×105×9.8 = 3.92×106N, A = 600m2. ΔP = 3.92×106600 = 6533.33Pa≈6533Pa. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 6533 Pa. This satisfies dimensional consistency and physical conditions given, so option C is scientifically correct.
Ref: NCBI Bookshelf, Ideal Gas Law: P1V1/T1 = P2V2/T2.
Viscous force: F = ηvAl. η = 0.02Pa s, v = 0.5m/s, A = 0.05m2, l = 0.2×10−3m. F = 0.02×0.5×0.050.2×10−3 = 0.02×125 = 2.5N. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 2.5 N. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.
Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.
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.
Total pressure: P = Pa+ρgh. Pa = 1.01×105Pa, ρ = 806kg/m3, g = 10m/s2, h = 3m. P = 1.01×105+806×10×3 = 1.01×105+24180 = 1.2518×105Pa. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 1.25 × 10⁵ Pa. This satisfies dimensional consistency and physical conditions given, so option C is scientifically correct.
Ref: NCBI Bookshelf, Ideal Gas Law: P1V1/T1 = P2V2/T2.
Pi = Pa+ρgh+2Sr. Pa = 1.01×105Pa, ρgh = 1000×9.8×0.15 = 1470Pa. 2Sr = 2×0.07273.5×10−3 = 41.54Pa. Pi = 1.01×105+1470+41.54 = 1.02471×105Pa. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 1.025 × 10⁵ Pa. This satisfies dimensional consistency and physical conditions given, so option C is scientifically correct.
Ref: NCBI Bookshelf, Ideal Gas Law: P1V1/T1 = P2V2/T2.
Excess pressure in a bubble: ΔP = 4Sr. S = 0.027N/m, r = 8×10−3m. ΔP = 4×0.0278×10−3 = 13.5Pa. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 13.5 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.
S = F2l, F = 1.2×10−2N, l = 0.25m. S = 1.2×10−22×0.25 = 0.024N/m. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 0.024 N/m. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.
Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.
Total pressure: P = Pa+ρgh. Pa = 1.01×105Pa, ρ = 1.03×103kg/m3, g = 9.8m/s2, h = 4m. P = 1.01×105+1.03×103×9.8×4 = 1.01×105+40376 = 1.41376×105Pa. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 1.41 × 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.
