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.

Answer

`(a)0,(b)h=0.25cm, (c ) a=g//6("upward")`

Explanation

(a) Liquid A is exerting force on the cylinder symmetrically from all sides. Therefore net force is zero.
(b) Weight = Buoyancy
mg=Buoyancy of liquid `A+` Buoyancy of liquid B
`(h+h_(A)+h_(B))Arhog=h_(A)rho_(A)Ag+h_(B)rho_(B)Ag`
`h=0.25cm`
(c ) `W-B=ma`
`[h_(A)Arho_(A)g+(h_(B)+h)Arho_(B)g]-(h+h_(A)+h_(B))Arhog=(h+h_(A)+h_(B))rhoA`
`hrho_(B)g=(h_(1)+h_(A)+h_(B))rho`
`0.25xx1.2xxg=(0.25+1.2+0.8)xx0.8xxa`
`0.25xx1.2xxg=2.25xx0.8a`
`a=(g)/(6)`