This is a problem of fluid statics in a uniformly accelerating (non-inertial) frame. When a container of liquid is given a constant acceleration, we analyse it by working in the frame of the container and adding a fictitious (d'Alembert) inertia force. The pressure variation with depth then depends on the effective gravitational acceleration felt by the fluid.

For the elevator accelerating upward with acceleration a, consider a small fluid column of depth h and cross-sectional area A. Newton's second law on that column (taking upward as positive) gives:

• Upward force from pressure below − weight − force to accelerate the column = mass × a
• P·A − P_atm·A − ρ·g·(A·h) = ρ·(A·h)·a
• Dividing by A: P − P_atm = ρgh + ρah = ρ(g + a)h

Therefore the pressure at depth h is:

P = P_atm + ρ(g + a)h

Physically, upward acceleration makes the fluid "feel heavier" — the effective gravity is (g + a) — so the pressure builds up faster with depth than under gravity alone. Useful limiting checks confirm the sign convention:

• If the elevator accelerated downward with a, effective gravity would be (g − a) and P = P_atm + ρ(g − a)h.
• In free fall (a = g downward), the pressure everywhere equals P_atm — the classic weightlessness result.

This is why P_atm + ρgh is wrong: it is the ordinary static result and omits the extra inertia term, so it underestimates the pressure. P_atm − ρ(g − a)h is wrong both in sign and because subtracting would imply pressure decreasing with depth, which is impossible for a liquid whose surface is at the top. P_atm − ρgh is likewise incorrect in sign. Only P_atm + ρ(g + a)h correctly reflects the increased effective gravity during upward acceleration.