9 public questions tagged with this topic.
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.
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.
Pressure increases with depth due to the weight of the fluid column above, as given by P = ρgh. The greater the depth, the more fluid mass contributes to the pressure at that point.
Ref: NCBI Bookshelf, Ideal Gas Law: P1V1/T1 = P2V2/T2.
The hydrostatic paradox demonstrates that pressure in a fluid at rest depends only on depth (ρgh) and not on the shape or volume of the container. At the same depth, the pressure is identical regardless of the vessel’s shape.
Ref: NCBI Bookshelf, Ideal Gas Law: P1V1/T1 = P2V2/T2.
ρgh represents the hydrostatic pressure due to the weight of the fluid column above a point, increasing with depth, as it accounts for gravitational potential energy per unit volume. As per NCERT, applying relevant law/formula with correct units and sign convention leads to Hydrostatic pressure. This satisfies dimensional consistency and physical conditions given, so option C is scientifically correct.
Ref: NCBI Bookshelf, Ideal Gas Law: P1V1/T1 = P2V2/T2.
P = Pa+ρgh. Pa = 1.01×105Pa, ρ = 1.03×103kg/m3, g = 9.8m/s2, h = 800m. P = 1.01×105+1.03×103×9.8×800 = 1.01×105+8.0784×106 = 8.1794×106Pa. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 8.18 × 10⁶ 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.
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.
Pi = Pa+ρgh+2Sr. Pa = 1.01×105Pa, ρgh = 1000×10×0.25 = 2500Pa. 2Sr = 2×0.07276×10−3 = 24.23Pa. Pi = 1.01×105+2500+24.23 = 1.03524×105Pa. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 1.035 × 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 = 1000×10×0.45 = 4500Pa. 2Sr = 2×0.07275.5×10−3 = 26.44Pa. Pi = 1.01×105+4500+26.44 = 1.05526×105Pa. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 1.055 × 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.
