A beaker containing a liquid of density rho m | vedclass

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(B) Consider an element of the liquid of height $dh$ and area of cross-section $A$ at a depth $h$ below the surface of the liquid. Let $p$ and $p+dp$

Vedclass · Accepted Answer

$h\rho (g+a)$

Vedclass · Answer

(B) Consider an element of the liquid of height $dh$ and area of cross-section $A$ at a depth $h$ below the surface of the liquid. Let $p$ and $p+dp$ be the liquid pressures at the upper and lower surfaces of the element, respectively.The mass of the liquid in the element is $dm = A \cdot dh \cdot \rho$.The net upward force on the element is given by $F_{net} = (p+dp)A - pA - dm \cdot g$.Since the element moves upward with an acceleration $a$, by Newton's second law, $F_{net} = dm \cdot a$.Therefore, $A \cdot dp - dm \cdot g = dm \cdot a$.$A \cdot dp = dm(g+a)$.Substituting $dm = A \cdot dh \cdot \rho$, we get $A \cdot dp = (A \cdot dh \cdot \rho)(g+a)$.$dp = \rho(g+a) dh$.Integrating both sides from the surface $(h=0, p=0)$ to depth $h$, we get $p = \int_0^h \rho(g+a) dh = \rho(g+a)h$.

$A$ 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

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