The top surface of an incompressible liquid is open to the atmosphere. The pressure at a depth h₁ below the surface is p₁∙ How does the pressure p₂ at depth h₂ = 2_div compare with p₁?

(A) p₂ > 2_p1 (B) P₂ = 2_p1 (C) P_{2 p1}

The pressure at a given depth in an incompressible liquid is given by the hydrostatic pressure equation:

p = p₀ + ρgh

where:

p is the pressure at depth h,
p₀ is the atmospheric pressure at the surface,
ρ is the density of the liquid,
g is the acceleration due to gravity,
h is the depth below the surface.
Given that at depth h₁, the pressure is:

p₁ = p₀ + ρgh₁

At depth h₂ = 2h₁, the pressure is:

p₂ = p₀ + ρg(2h₁)

Expanding this:

p₂ = p₀ + 2ρgh₁

Since p₁ = p₀ + ρgh₁, we substitute this into the equation for p₂:

p₂ = p₀ + 2ρgh₁ = 2(p₀ + ρgh₁) - p₀
= 2p₁ - p₀

Since atmospheric pressure p₀ is a positive value, we see that:

p₂ < 2p₁

Thus, the correct answer is:

(C) p₂ < 2p₁