A beaker containing a liquid of density $\rho $ moves up with an acceleration $a$ ; the pressure due to the liquid at a depth $h$ below the free surface of the liquid is
Diffcult
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Consider an element of the liquid of height $dh$ and area of cross$-$section $\mathrm{A}$ at a depth $\mathrm{h}$ below the surface of the liquid. Let $p$ and $p$ $+$ $\mathrm{dp}$ be the liquid pressures at the upper and lower surfaces of the element. Mass of the liquid in the element, $\mathrm{dm}=$ $Adh$$\rho$
Net upward force on the element
$=[(p+d p) A-p A]-g d m$
Or $\mathrm{A} \mathrm{dp}-\mathrm{g} \mathrm{dm}=\mathrm{a} \mathrm{dm}$
(As the element moves up with an acceleration a)
Or $\mathrm{A} \mathrm{dp}=(\mathrm{g}+\mathrm{a}) \mathrm{dm}=(\mathrm{g}+\mathrm{a})$ $Adh\rho.$
$\int \mathrm{d} \mathrm{p}=\int \rho(\mathrm{g}+\mathrm{a}) \mathrm{dh}$ Or $\mathrm{p}=\rho(\mathrm{g}+\mathrm{a}) \mathrm{h}$
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