A beaker containing a liquid of density ρ moves up with an acceleration ' a '. The pressure due to the liquid at a depth h below free surface of the liquid is

Question

A beaker containing a liquid of density ρ moves up with an acceleration ' a '. The pressure due to the liquid at a depth h below free surface of the liquid is (A) h ρ g (B) h ρ(g-a) (C) h ρ(g+ a) (D) 2 h ρ g((g+ a/g-a))

Tardigrade Admin · Accepted Answer

Due to upward acceleration pseudo force will act downwards so value of acceleration due to gravity will increase by 'a ' ∴ g'=(g +a) P=ρ g 'h ⇒ P=ρ(g +a) h (Substitute g' )

Q. A beaker containing a liquid of density $ρ$ moves up with an acceleration ' $a$ '. The pressure due to the liquid at a depth $h$ below free surface of the liquid is

$h ρ g$

$h ρ (g - a)$

$h ρ (g + a)$

$2 h ρ g (\frac{g + a}{g - a})$

Due to upward acceleration pseudo force will act downwards so value of acceleration due to gravity will increase by ' $a$ '
$∴ g^{'} = (g + a)$
$P = ρ g^{'} h$
$\Rightarrow P = ρ (g + a) h$
(Substitute $g^{'}$ )

Q. A beaker containing a liquid of density ρ moves up with an acceleration ' a '. The pressure due to the liquid at a depth h below free surface of the liquid is

hρg

hρ(g−a)

hρ(g+a)

2hρg(g−ag+a​)