A container of a large uniform cross-sectional area `A` resting on a horizontal surface holds two immiscible, non viscous and incompressible liquids of densities `d` and `2d`, each of height `H//2` as shown in figure. The lower density liquid is open to atmosphere. A homogeneous solid cylinder of length `L(L lt (H)/(2))`, cross-sectional area `A//5` is immersed such that it floats with its axis vertical of the liquid-liquid interface with length `L//4` denser liquid. Determine
(a) density `D` of the solid and
(b) the total pressure at the bottom of the container. (Atmospheric pressure `= P_0`).

.

Answer

(i) (a) `(5d)/(4), (b) `P_0 + (dg(6H + L))/(4) `

Explanation

(i) (a) `(5 d ) /(4)`, (b) ` P_0 + (dg (6 H + L))/(4) `
(i) (a) Considering vertical equilibrium of cylinder weight of cylinder = upthrust due to upper liquid + upthrust due to lower liquid
` therefore ((A)/(5)) (L) D . G = ((A)/(5)) ((3L)/(4)) (d) g + ((A)/(5)) ((L)/(4)) (2d ) (g) `
` therefore D = ((3)/(4)) d + ((1)/(4)) (2d) rArr D = (5)/(4) d `
(b) Considering vertical equilibrium of two liquids and the cylinder.
` (p - p _ 0 ) A ` = weight of two liquids + weight of cylinder
` therefore P = P_ 0 + ("Weight of liquids + Weight of cylinder")/(A) " " ...(i) `
Now, weight of cylinder ` = ((4)/(5)) (L) (D) (g) = ((A)/(5) Lg ) ((5)/(4) d ) = (ALd g )/(4) `
weight of upper liquid ` = ((H)/(2) Ad g ) ` and weight of lower liquid ` = (H)/(2) A (2d ) g = HA gh `
` therefore ` Total weight of two liquids ` = (3)/(2) HA dg `
` therefore ` From equation (i) pressure at the bottom of the container will be
` therefore p = p _ 0 + ( ((3)/(2)) HA dg + (ALdg )/(4))/ (A) or p = p _ 0 + (dg (6H + L))/(4 ) `