Question

A wooden block is placed in a tub of liquid with density rho_1 and we observe that exactly half the volume of the block is submerged in the liquid. We slowly pour a second liquid (immiscible with the first liquid) with density rho_2 on top of the first liquid until the interface between the top liquid and the air coincides with the top surface of the block. Show that x, the fraction of the block's volume submerged in the lower liquid (x = V_sub/V_total) is x = rho_1 - 2 rho_2/2(rho_1 - rho_2)

Answer 1

Here ,

let us assume that the density of wooden block is p

Now , as the wooden block is half submerged in the first liquid

relation between density of first liquid and woooden block

p1 = 2*p

p = p1/2

NOw , let the volume of the block is V

for the fraction is x

after pouring the second liquid

Using Archimedes principle

p1 * x * V * g + p2 * (1 - x)*V * g = p * V * g

p1 * x + p2 - p2 * x = p1/2

2 p1 * x + 2 p2 - 2 p2 * x = p1

x = (p1 - 2p2)/(2 *(p1 - p2))

the fraction of the block in the lower liquid is (p1 - 2p2)/(2 *(p1 - p2))