A container having an ideal liquid of density $ρ$ is moving with a constant acceleration of $\vec{a} = a_{x} \hat{i} + a_{z} \hat{k}$ where x direction is horizontal and z is vertically upward. The container is open at the top. In a reference frame attached to the container with origin at bottom corner (see figure), write the pressure at a point inside the liquid at co-ordinates (x, y, z). The pressure is P₀ at origin.

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$P_{0} - ρ a_{x} x + ρ (g - a_{z}) z$

$P_{0} + ρ a_{x} x - ρ g z$

$P_{0} - ρ a_{x} x - ρ (g + a_{z}) z$

$P_{0} - ρ a_{x} x - ρ a_{z} z$

answer is C.

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Detailed Solution

$\frac{\partial P}{\partial x} = - {ρa}_{x}, \frac{\partial P}{\partial y} = 0 and \frac{\partial P}{\partial z} = - ρ (g + a_{z})$
Pressure is independent of y
Total differential of pressure P will be
$\begin{array}{l} dP = (\frac{\partial P}{\partial x}) \cdot dx + (\frac{\partial P}{\partial z}) dz = - {ρa}_{x} dx - ρ (g + a_{z}) dz \\ ∴ \int_{P_{0}}^{P} dP = - {ρa}_{x} \int_{0}^{x} dx - ρ (g + a_{z}) \int_{0}^{z} dz \\ P - P_{o} = - {ρa}_{x} x - ρ (g + a_{z}) z \\ ∴ P = P_{o} - {ρa}_{x} \cdot x - ρ (g + a_{z}) z \end{array}$