1) Consider a vessel containing a fluid. Let the acceleration of vessel in vertical direction be, a_{y}. Consider a small cylindrical element dy at a height y from bottomFrom figure,

P₁ A – P₂A – mg = may

P₁ A – P₂A = m (g + ay)

But (P₁ – P₂) A2 = A dy (g + ay)

– dp/ dy = – (g + ay)

2) Consider a vessel containing a fluid. Let the acceleration of vessel in horizontal direction beFrom diagram

(P₁ – P₂) A = Adx (ax)

Δp/Δx = – ax

If θ is the angle made by the free liquid surface with horizontal, then

tanθ = dh/dx

But we have

(P₁ – P₂) A = Adx (ax)

(h₁ – h₂) gA = Adx (ax)

dh/ dx = aₓ/g

tanθ = aₓ/g