38 public questions tagged with this topic.
ΔP = ρgh. ρ = 1.06×103kg/m3, g = 10m/s2, h = 0.18m. ΔP = 1.06×103×10×0.18 = 1908Pa. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 1908 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.
Gauge pressure: Pg = ρgh = 1.03×103×10×300 = 3.09×106Pa. F = PgA = 3.09×106×0.08 = 2.472×105N. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 2.47 × 10⁵ N. 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 equation: P1+12ρv12 = P2+12ρv22. P1 = 1.7×105Pa, v1 = 3.5m/s, v2 = 6m/s, ρ = 1000kg/m3. P2 = 1.7×105+12×1000(12.25−36) = 1.7×105−11875 = 1.58125×105Pa. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 1.58 × 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.
Total pressure: P = Pa+ρgh. Pa = 1.01×105Pa, ρ = 1.03×103kg/m3, g = 9.8m/s2, h = 3.4m. P = 1.01×105+1.03×103×9.8×3.4 = 1.01×105+34319.6 = 1.353196×105Pa. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 1.35 × 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.
Pa = ρgh. ρ = 13.6×103kg/m3, g = 10m/s2, h = 0.75m. Pa = 13.6×103×10×0.75 = 1.02×105Pa. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 1.02 × 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.
Gauge pressure: Pg = ρgh. ρ = 1.06×103kg/m3, g = 9.8m/s2, h = 1.5m. Pg = 1.06×103×9.8×1.5 = 15582Pa. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 1.56 × 10⁴ 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.
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.
Pascal’s law: P = F1A1 = F2A2. F2 = F1×A2A1. F1 = 120N, A1 = 0.008m2, A2 = 0.032m2. F2 = 120×0.0320.008 = 120×4 = 480N. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 480 N. This satisfies dimensional consistency and physical conditions given, so option C is scientifically correct.
Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.
Pi = Pa+ρgh+2Sr. Pa = 1.01×105Pa, ρgh = 1.2×103×9.8×0.5 = 5880Pa. 2Sr = 2×0.0254×10−3 = 12.5Pa. Pi = 1.01×105+5880+12.5 = 1.06892×105Pa. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 1.069 × 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.
In a fluid at rest, any tangential (shear) force would cause the fluid to flow due to its inability to resist shear stress. According to Newton’s third law, if a tangential force existed, the fluid would exert an equal and opposite force, leading to motion. Since the fluid is at rest, no such motion occurs, and the force must be normal to the surface.
Ref: NCBI Bookshelf, Ideal Gas Law: P1V1/T1 = P2V2/T2.
Gauge pressure: Pg = ρgh = 1.03×103×10×600 = 6.18×106Pa. F = PgA = 6.18×106×0.05 = 3.09×105N. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 3.09 × 10⁵ N. This satisfies dimensional consistency and physical conditions given, so option C is scientifically correct.
Ref: NCBI Bookshelf, Ideal Gas Law: P1V1/T1 = P2V2/T2.
Bernoulli’s equation: P1+12ρv12+ρgh1 = P2+12ρv22+ρgh2. P1 = 1.2×105, v1 = 2m/s, h1 = 1m, v2 = 4m/s, h2 = 0m. Left: 1.2×105+12×1000×4+1000×10×1 = 1.2×105+2000+10000 = 1.32×105. Right: P2+12×1000×16 = P2+8000. 1.32×105 = P2+8000, P2 = 1.24×105Pa.
Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.
