A uniform solid cylinder of density `0.8g//cm^3` floats in equilibrium in a combination of two non-mixing liquids A and B with its axis vertical.
The densities of the liquids A and B are `0.7g//cm^3` and `1.2g//cm^3`, respectively. The height of liquid A is `h_A=1.2cm.` The length of the part of the cylinder immersed in liquid B is `h_B=0.8cm`.

(a) Find the total force exerted by liquid A on the cylinder.
(b) Find h, the length of the part of the cylinder in air.
(c) The cylinder is depressed in such a way that its top surface is just below the upper surface of liquid A and is then released. Find the acceleration of the cylinder immediately after it is released.

Explanation

(a) In this problem, liquid A exerts only horinzontal force on all area elements of the cylinder in contact with it. All these force will cancel out, due to symmetry. Thus the total force exerted by the liquid A on the cylinder is zero. It is to be noted that the buoyant force acts on the horizontal bottom surface of cylinder only due to liquid B.
(b) Let a be the uniform area of cross-section of the cylinder. weight of cylinder,
`W = (h+h_(A)+h_(b)) xx a xx 0.8g`
`=(h+1.2 +0.8) xx a xx 0.8 g` ltbr. `=(h+2) a xx 0.8 g`
The buoyant force on the cylinder due to liquids
= weight of liquid A displaced +weight of liquid B displaced
`=ah_(A)rho_(A)g + ah_(B)rho_(B)g`
`=a(1.2 xx 0.7 + 0.8 xx 1.2)g = 1.8 ag`
As the cylinder is in equilibrium , so
weight of cylinder = buoyandt force due to liquids
i.e, `(h+2)a xx 0.8g = 1.8ag`
or `(h+2) xx 0.8 - 1.8`
or `h=(1.8)/(0.8) - 2 = 0.25 cm`
(c ) On depressing the cylinder further so as to just submerge in liquid , additional volume of the liquid `B (=0.25a)` is displaced. this produces an unbalanced effective upward force of bouyancy which will be responsible for the upward ac celeration of the cylinder.
The unbalanced effective upward force of buoyancy is `F = 0.25 a rho_(B)g = 0.25 a xx 1.2 xx g = 0.3 ag`
The mass of the cylinder
`m=(h+h_(A)+h_(B)) a xx 0.8`
`=(0.25 + 1.2 + 0.8) a xx 0.8 = 1.8 a`
Upward ac celeration of cylinder
`=F/m = (0.3 ag)/(1.8a) = g/6`.