Two liquids of densities ρ1 and ρ2(ρ2=2ρ1) are filled up behind a square wall of side 10 m as shown in figure. Each liquid has a height of 5 m. The ratio of the forces due to these liquids exerted on upper part MN to that at the lower part NO is (Assume that the liquids are not mixing):

Question

Two liquids of densities ρ1 and ρ2(ρ2=2ρ1) are filled up behind a square wall of side 10 m as shown in figure. Each liquid has a height of 5 m. The ratio of the forces due to these liquids exerted on upper part MN to that at the lower part NO is (Assume that the liquids are not mixing): (A) 1/2 (B) 2/3 (C) 1/4 (D) 1/3

Tardigrade Admin · Accepted Answer

F1=(ρ gh/2)× A F2(ρ gh+(2ρ gh/2))A (F1/F2)=(1/4)

$1/2$

$2/3$

$1/4$

$1/3$

$F_{1} = \frac{ρ g h}{2} \times A$
$F_{2} (ρ g h + \frac{2 ρ g h}{2}) A$
$\frac{F _{1}}{F _{2}} = \frac{1}{4}$

1/2

2/3

1/4

1/3

F1​=2ρgh​×A F2​(ρgh+22ρgh​)A F2​F1​​=41​

$1/2$

$2/3$

$1/4$

$1/3$

$F_{1} = \frac{ρ g h}{2} \times A$
$F_{2} (ρ g h + \frac{2 ρ g h}{2}) A$
$\frac{F _{1}}{F _{2}} = \frac{1}{4}$