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Pressure at depth h below the free surface of a liquid of density ρ (with atmospheric pressure P₀ above) is:
AP₀ + ρ g h
BP₀ − ρ g h
Cρ g h
DP₀ · ρ g h
Answer & Solution
Correct answer: A. P₀ + ρ g h
1. Pressure increases linearly with depth: dP/dh = ρ g.
2. Integrating from the free surface to depth h gives P(h) − P₀ = ρ g h.
3. So P(h) = P₀ + ρ g h.
4. Pressure depends on depth h, not on the shape of the container.
_Source: NCERT Class 11 Physics Ch 9 §9.3 Pressure due to a Fluid Column_
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