What is the Velocity of the Larger Cube as the Smaller Cube Melts?

[geniusno.198]

Homework Statement

Two cubes made of the same material and edge length 'a' and '2a'. They are placed in a liquid of density σ such that the larger cube is completely submerged when the smaller cube is placed on top of it and the lower surface of smaller cube is just touching the surface of the liquid.
Now, if the smaller cube is heated sucjh that it melts uniformly and very slowly (reducing all dimensions simultaneously) such that the edge reduces at a constant rate α then the velocity of the larger cube as a function of time will be (assuming molten material evaporates immediately).

Homework Equations

let the densities of both the blocks be 'ρ'
Mass of the larger block= 8ρa^3 (mass=volume*density)
Mass of smaller block= ρa^3
since weight of both the blocks is balanced by the buoyant force the weight of both the blocks together is equal to the buoyant force.
(ρa^3+8ρa^3)g = Volume of body submerged*density of liquid*g
=8ρa^38σg
therefore ρ=8σ/9

The Attempt at a Solution

I thought what we basically have to find is the rate at which the mass of the small block is changing. so we we have to find the differential of the mass of the smaller block with respect to time. therefore d(8σa^3/9)/dt = 8σa^2/3 8 da/dt. but according to the question da/dt=α. I got stuck here,I don't know what I should do next.

 

[geniusno.198]

I am really sorry, but could anyone please help me ?

 

[ehild]

The problem might mean that the evaporation rate is so slow that the blocks are almost in equilibrium all the time, that is the weight and buoyant force are balanced. Find the immersion depth of the big cube at time t when the side length of the small block is a-αt. In this case, the velocity is the time derivative of the immersion depth.

ehild