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

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Solution

Net upward force=extra upthrust
$= s h ρ_{B} g$
$∴$ Net acceleration $a = \frac{f o r c e}{m a s s o f c y l i n d e r}$
or $a = \frac{s h ρ_{B} g}{s (h + h_{A} + h_{B})} ρ_{c y l i n d e r}$
or $a = \frac{h ρ_{B} g}{h + h_{A} + h_{B}} ρ_{c y l i n d e r}$
Substituting the values of h, $h_{A}$ , $h_{B}$ , $ρ_{B}$ and $ρ_{c y l i n d e r}$ we get
$a = \frac{g}{6}$ (upwards)

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A uniform solid cylinder of density $0.8 g / {cm}^{3}$ 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 $0.7 g / {cm}^{3}$ and $1.2 g / {cm}^{3}$ respectively. The height of liquid A is $h_{A} = 1.2 cm$ . The length of the part of the cylinder immersed in liquid B is $h_{B} = 0.8 cm$ .