A uniform solid cylinder of density $0.8 g / c m^{3}$ 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 $0.7 g / c m^{3}$ and $1.2 g / c m^{3}$ , respectively. The height of liquid A is $h_{A} = 1.2 c m$ . The length of the part of the cylinder with liquid B is $h_{B} = 0.8 c m$ and $h = .2 c m$ . 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. The acceleration of the cylinder immediately just after it is released is given as $\frac{x}{3} m / s^{2}$ . Find $x$ :

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Solution

Mass of cylinder $= (h + h_{A} + h_{B}) A ρ$
$= (0.2 + 1.2 + 0.8) A \times 0.8 = 1.8 A$
When the cylinder is depressed just completely, its height h goes in liquid B, so extra buoyant force due to liquid B acts on it. Due to this force, cylinder is accelerated upward.
Extra buoyant force $F = (h_{A}) ρ_{B} g$
$= 0.25 A \times 1.2 g = 0.3 A g$
Acceleration is given by $a = \frac{F}{m} = \frac{0.3 A g}{1.8 A} = \frac{g}{6}$
$= \frac{10}{6} = \frac{5}{3} m / s^{2}$

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