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#seawater depth

3 public questions tagged with this topic.

Fluid Pressure, Pascal's Law and Manometers

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.

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Bernoulli's Theorem and Applications

What is the absolute pressure at 250m depth in seawater (ρ\=1.03×103kg/m3) with Pa\=1.01×105Pa? (Take g\=9.8m/s2)

P = Pa+ρgh. Pa = 1.01×105Pa, ρ = 1.03×103kg/m3, g = 9.8m/s2, h = 250m. P = 1.01×105+1.03×103×9.8×250 = 1.01×105+2.5235×106 = 2.6245×106Pa. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 2.6 × 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.

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Equation of Continuity and Flow Rate

What is the absolute pressure at 450m depth in seawater (ρ\=1.03×103kg/m3) with Pa\=1.01×105Pa? (Take g\=9.8m/s2)

P = Pa+ρgh. Pa = 1.01×105Pa, ρ = 1.03×103kg/m3, g = 9.8m/s2, h = 450m. P = 1.01×105+1.03×103×9.8×450 = 1.01×105+4.5411×106 = 4.6421×106Pa. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 4.6 × 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.

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