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 (in `m//s^(2)`) 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//s^(2))`

Explanation

Let the volume of the cylinder be `V`
When the cylinder is floating
upthrust =weight.
`rho((3)/(4)V)g=mg`, `V=(4m)/(3rho)`
Let the acceleration of the particle vessel be `A`(upwards)
In the reference frame of vessel, the acceleration of the cylinder is `A//3`.
`:. mg+mA-unthrust=m(A//3)`
`mg+mA-rhoVg'=m(A//3)`
[Where `g'=g+A=` effective value of `g` for upthrust]
`mg+mA-rhoV(g+A)=m(A//3)`
`mg-(4)/(3)m(g+A)= -m(2)/(3)A`
`A=(-g//2)` upwards
The acceleration of the vessel should be `g//2` downwards.