A closed rectangular vessel completely filled with a liquid of density ρ moves with an acceleration a=g. The value of the pressure difference at point A and B i.e., (P1-P2) is

Question

A closed rectangular vessel completely filled with a liquid of density ρ moves with an acceleration a=g. The value of the pressure difference at point A and B i.e., (P1-P2) is (A) ρ g(b-h) (B) ρ g h / 2 (C) ρ g h (D) (ρ g(b+h)/2)

Tardigrade Admin · Accepted Answer

Let pressure at the opening be P0. Then, P1=P0+ρ g b while P2=P0+ρ g h . <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/01520674baa60dd66ce7a03ce35b9670-.png" /> Subtracting, P1-P2=ρ g(b-h)

Q. A closed rectangular vessel completely filled with a liquid of density $ρ$ moves with an acceleration $a = g$ . The value of the pressure difference at point $A$ and $B$ i.e., $(P_{1} - P_{2})$ is

$ρ g (b - h)$

$ρ g h /2$

$ρ g h$

$\frac{ρ g ( b + h )}{2}$

Let pressure at the opening be $P_{0}$ . Then,
$P_{1} = P_{0} + ρ g b$ while $P_{2} = P_{0} + ρ g h .$

Subtracting, $P_{1} - P_{2} = ρ g (b - h)$

Q. A closed rectangular vessel completely filled with a liquid of density ρ moves with an acceleration a=g. The value of the pressure difference at point A and B i.e., (P1​−P2​) is

ρg(b−h)

ρgh/2

ρgh

2ρg(b+h)​