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#submarine window

3 public questions tagged with this topic.

What is the force on a submarine window (0.06m2) at 400m depth in seawater (ρ\=1.03×103kg/m3), interior at atmospheric p

Gauge pressure: Pg = ρgh = 1.03×103×9.8×400 = 4.0376×106Pa. F = PgA = 4.0376×106×0.06 = 2.42256×105N. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 2.4 × 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.

What is the force on a submarine window (0.05m2) at 600m depth in seawater (ρ\=1.03×103kg/m3), interior at atmospheric p

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.

What is the force on a submarine window (0.07m2) at 350m depth in seawater (ρ\=1.03×103kg/m3), interior at atmospheric p

Gauge pressure: Pg = ρgh = 1.03×103×10×350 = 3.605×106Pa. F = PgA = 3.605×106×0.07 = 2.5235×105N. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 2.5 × 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.