Question

A cubical block of density d is floating on water surface with fraction x submergerged in water it is kept in lift moving downward

Answer 1

Explanation:

Apply Archimedes principle when system is at rest,

weight of the block = Buoyant force by liquid

\rhob V g = \rhoL V/2 g

so

\rhob/\rhoL =1/2

Now, when system accelerates upward with g/3 draw FBD of the block, Buoyant force upwards and weight downwards

Being part of the system block also goes up with g/3,

making equation from 2nd law of motion

Fb - Mg = Mg/3

Fb = 4Mg/3

\rholiquid Vvolume of block immersed g = 4/3 \rhoblock Vtotal volume of the block g

hence

Vvolume of block immersed / Vtotal volume of the block = 2/3