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

Liquid A exerts all round horizontal force on the cylinder and so net on it is zero.
If A is the area of cross – section of the cylinder, then for the floating cylinder 
Weight of the cylinder = buoyant force by the liquids 
 $(h_A+h_B+h)A\times 0.8g=\left(h_{A}A\right){\rho }_{A}g+\left(h_{B}A\right){\rho }_{B}g$
 $(1.2+0.8+h)A\times 0.8g=\left(1.2A\right)\times 0.7g+\left(0.8A\right)\times 1.2g$
$\therefore h=0.25cm$ 
The extra buoyant force 
 $F_h=\left(Ah\right){\rho }_{B}g$                
Acceleration  $a=\frac{F_h}{m}$
                         $=\frac{Ah{\rho }_{B}g}{m}$  
Here   $m=(h_A+h_B+h)A\times 0.8$             
         $=(1.2+0.8+0.25)A\times 0.8$
          $=1.8A$
$\therefore a=\frac{A\times 0.25\times 1.2g}{1.8A}$ 
                  $=\frac{g}{6}m/s^2$ .