Question

A container of large uniform cross-sectional area A resting on a horizontal surface holds 2 immiscible liquids of densities d and 2d ,each of height h/2 . A homogenous solid cylinder of lenght of L (L

Answer 1

The cylinder floats such that it is L/4 parts is in the dense liquid
3L/4 is in the rearer liquid

weight of cylinder = upthrust due to liquid
density of solid= D
density of upper liquid= d
density of lower liquid= 2d
area= A/5

weight= D x g x V
W= ALDg/5

upthrust due to upper liquid= volume dipped x d xg
F1= 3ALdg/20

upthrust due to lower liquid= volume dipped x 2d x g
F2= 2ALdg/20

Prnciple of floatation = W= F1 + F2
by solving the equation we will get D= 5d/4