Question

I place a block in a tub of mercury, and find that 1/4 of its volume is submerged.
Now I pour just enough water so that the object is fully submerged. At the end of this
procedure, approximately how much of the block’s volume will remain immersed in
mercury? The density of mercury and water are, respectively, 13.6 and 1.0 g/cm3

Answer 1

• If a block is submerged in mercury, we can write an equation as below.

weight of the block = upthrust force by mercury

$\\ mg = U$

$mg = Vdg$

$mg = (1/4)V X 13.6 X g$

$m = 13.6V/4$

(V = volume of the block , m= mass of the block)

• If a block is submerged in water & mercury,

weight of the block = upthrust force by mercury + upthrust force by mercury

mg = U1 +U2

(13.6V/4) = (V' X 13.6 X g) + (V-V') x 1 x g

13.6V/4 = 13.6V' + V -V'

9.6V/4 = 12.6V'

$V' = 0.19V$

$V' = (19/100)V$

submerged volume in mercury = (19/100) of block's volume