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Consider an open cylindrical vessel containing a fluid of density $\rho$. The pressure at its open surface is atmospheric pressure $P_\text{atm}$. The pressure at a depth $y$ below the top surface is given by \begin{align} P=P_\text{atm}+\rho g y. \end{align} The hydrostatic pressure increases linearly with depth $y$.

The pressure difference between two horozontal layes separated by a distance $y$ is given by \begin{align} \Delta P=\rho g y. \end{align}

**Problem (IIT JEE 1999):**
A closed compartment containing gas is moving with some acceleration in horizontal direction. Neglect effect of gravity. Then, the pressure in the compartment is,

- same everywhere
- lower in front side
- lower in rear side
- lower in upper side

**Solution:**
Let $A$ be the cross-section area and $\rho$ be the gas density. Consider a small element of width $\mathrm{d}x$ at a distance $x$ (see figure). Let pressure at the distance $x$ be $p$ and that at distance $x+\mathrm{d}x$ be $p+\mathrm{d}p$. The net force on the element due to pressure difference is,
\begin{align}
\mathrm{d}F & =[pA-(p+\mathrm{d}p)A] \\
&=-A\mathrm{d}p,\qquad \text{(towards right)}.
\end{align}

This force provides acceleration $a$ to the element. Apply Newton's second law on the element, \begin{align} -A\mathrm{d}p=ma=(\rho A \mathrm{d}x) a, \end{align} to get, \begin{align} {\mathrm{d}p}/{\mathrm{d}x}=-\rho a. \end{align} Negative sign indicates that the pressure decreases with $x$ i.e., as we go from the left to the right.