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#hydrostatics

7 public questions tagged with this topic.

What is the force on a submarine window (0.09m2) at 200m depth in seawater (ρ\=1.03×103kg/m3), interior at atmospheric p

Gauge pressure: Pg = ρgh = 1.03×103×9.8×200 = 2.0194×106Pa. F = PgA = 2.0194×106×0.09 = 1.81746×105N. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 1.82 × 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.

A manometer with mercury (ρ\=13.6×103kg/m3) shows a height difference of 0.25m. What is the pressure difference? (Take g

ΔP = ρgh. ρ = 13.6×103kg/m3, g = 9.8m/s2, h = 0.25m. ΔP = 13.6×103×9.8×0.25 = 33320Pa. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 3.33 × 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.

What is the gauge pressure at 2.5m depth in a tank of mercury (ρ\=13.6×103kg/m3)? (Take g\=10m/s2)

Gauge pressure: Pg = ρgh. ρ = 13.6×103kg/m3, g = 10m/s2, h = 2.5m. Pg = 13.6×103×10×2.5 = 3.4×105Pa. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 3.4 × 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.

What is the absolute pressure at 300m depth in seawater (ρ\=1.03×103kg/m3) with Pa\=1.01×105Pa? (Take g\=10m/s2)

P = Pa+ρgh. Pa = 1.01×105Pa, ρ = 1.03×103kg/m3, g = 10m/s2, h = 300m. P = 1.01×105+1.03×103×10×300 = 1.01×105+3.09×106 = 3.191×106Pa. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 3.19 × 10⁶ 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.