A vessel containing water of depth h is accelerated upward with an acceleration of g/2. The pressure at the bottom of the vessel is?

Asked in SSC JE 2016

Detailed Explanation

When a fluid container is subjected to linear vertical acceleration, the fluid experiences an effective gravitational acceleration according to d'Alembert's principle of dynamic equilibrium.

1. Consider a vertical fluid column of height \(h\) and cross-sectional area \(A\) accelerating vertically upward with acceleration \(a = g/2\):

2. Apply Newton's second law (\(\sum F_z = m \cdot a\)) to the column:

\[ P \cdot A - p_{\text{atm}} \cdot A - W = m \cdot a \]

3. Gauge pressure (\(p_{\text{atm}} = 0\)) with weight \(W = \rho g h A\) and mass \(m = \rho h A\):

\[ P \cdot A - \rho g h A = (\rho h A) \cdot a \]

\[ P = \rho (g + a) h \]

4. Substitute upward acceleration \(a = +\frac{g}{2}\):

\[ P = \rho \left(g + \frac{g}{2}\right) h = \rho \left(\frac{3g}{2}\right) h = \frac{3}{2} \rho g h = \frac{3\gamma h}{2} \]

(where \(\gamma = \rho g\) is the standard specific weight of water).

Why “3γh/2” Is Correct

An upward acceleration of \(g/2\) creates an additional downward inertial force equal to half the fluid's static weight. The total pressure at the bottom is therefore \(1.5\) times the static hydrostatic pressure, giving \(\frac{3\gamma h}{2}\).

Why the Other Options Are Wrong

• A. γh: γh is the standard static pressure when the vessel is at rest or moving at constant velocity (a = 0).
• B. γh/2: γh/2 corresponds to a vessel accelerating downward at a = g/2: P = ρ(g − g/2)h = γh/2.
• C. 2γh: 2γh corresponds to an upward acceleration of a = g (P = ρ(g + g)h = 2γh).

Key Concepts for Students

• Upward acceleration: P = ρ(g + a)h = γh(1 + a/g).
• Downward acceleration: P = ρ(g − a)h = γh(1 − a/g).
• Free fall (a = g downwards): P = ρ(g − g)h = 0 (weightlessness, pressure throughout becomes uniform/zero).

Practice more: this is question 25 in Hydrostatic Pressure MCQs – Part 2. Try the whole set and check every answer instantly.