A solid cylinder of height h and mass m is floating in a liquid of density ρ as shown in the figure. Find the acceleration of the vessel (in m/s2) 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. (Take g = 10 m/s2)

Let the volume of the cylinder be V When the cylinder is floating, upthrust weight. Hence,ρ34Vg=mg⇒V=4m3ρLet the acceleration of the particle vessel be A (upwards). In the reference frame of the vessel, the acceleration of the cylinder is A.3.∴ mg+mA− uptrust =mA3         mg+mA−ρVg′=mA3where g' = g + A = effective value of g for upthrust.∴     mg+mA−ρV(g+A)=mA3⇒     mg−43m(g+A)=−m23A    A=−g2 upwards The acceleration of the vessel should be g2=102=5ms−2 (downwards).