A solid cylinder of height h and mass m is floating in a liquid of density $\rho$ as shown in the figure. Find the acceleration of the vessel $\left(\text{ in }\mathrm{m}/{\mathrm{s}}^2\right)$  containing liquid for which the relative downward acceleration of the completely immersed cylinder w.r.t. vessel becomes equal to one-third of that of the vessel. $\left(\text{ Take }=\mathrm{g}=10\mathrm{m}/{\mathrm{s}}^2\right)$ 

Let the volume of the cylinder be V. When the cylinder is floating, upthrust=weight. Hence,

$\rho \left(\frac{3}{4}\mathrm{V}\right)\mathrm{g}=\mathrm{mg}\Rightarrow \mathrm{V}=\frac{4\mathrm{m}}{3\rho }$

Let the acceleration of the particle vessel be A (upwards). In the reference frame of the vessel, the acceleration of the cylinder is $\frac{{\displaystyle \mathrm{A}}}{3}$

$\therefore \mathrm{mg}+\mathrm{mA}-\text{ upthrust }=\mathrm{m}\left(\frac{\mathrm{A}}{3}\right)$

$\mathrm{mg}+\mathrm{mA}-{\mathrm{\rho Vg}}^1=\mathrm{m}\left(\frac{\mathrm{A}}{3}\right)$

Where ${\mathrm{g}}^1=\mathrm{g}+\mathrm{A}=\text{ effective }$ value of g for upthrust

$\therefore \mathrm{mg}+\mathrm{mA}-\rho \mathrm{V}(\mathrm{g}+\mathrm{A})=\mathrm{m}\left(\frac{\mathrm{A}}{3}\right)$

$\Rightarrow \mathrm{mg}-\frac{4}{3}\mathrm{m}(\mathrm{g}+\mathrm{A})=-\mathrm{m}\frac{2}{3}\mathrm{A}$

The acceleration of the vessel should be $\frac{\mathrm{g}}{2}=\frac{10}{2}=5{\mathrm{ms}}^{-2}\left(\text{ downward}\right)$