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))`

Answer

5

Explanation

Let the volume of the cylinder be `V`. When the cylinder is floating, upthrust`=`weight. Hence
`rho=(3/4V)g=mgimpliesV=(4m)/(3rho)`
Let 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-upthrust=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)`
`implies mg-4/3mg(g+A)=-m2/3A`
`A=(-g/2)`upwards
The acceleration of the vessel should be `g/=10/2=5ms^(-2)` (downwards)