12. 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.
a. 1/2
b. 1/3
c. 1/4
d. 1/5

Block submerged 1/4 volume in mercury has density 3.4 g/cm³, calculated from Archimedes’ principle where weight equals buoyant force from mercury. [execute_python] Adding water until fully submerged balances buoyant forces from mercury (fraction x) and water (1-x), yielding x ≈ 0.19 or 1/5 volume in mercury.​

Initial Equilibrium

Weight of block equals buoyant force from mercury: ρ_block × V × g = 13.6 × (V/4) × g. Thus, ρ_block = 13.6/4 = 3.4 g/cm³.​

Final Equilibrium

With water added, ρ_block × V × g = 13.6 × (xV) × g + 1.0 × ((1-x)V) × g. Substituting ρ_block = 3.4 gives x = (3.4 – 1)/(13.6 – 1) ≈ 0.1905. [execute_python]​

Option Analysis

• a. 1/2 (0.5): Too high; implies ρ_block ≈ 7.8, exceeding initial 3.4. Wrong.

• b. 1/3 (≈0.333): Still high; gives ρ_block ≈ 5.13. Wrong.

• c. 1/4 (0.25): Matches initial, ignores added water buoyancy. Wrong.

• d. 1/5 (0.2): Closest to 0.19; correct approximation. [execute_python]

A block placed in mercury submerges 1/4 of its volume due to buoyancy equilibrium, revealing density 3.4 g/cm³ relative to mercury’s 13.6 g/cm³. Pouring water until fully submerged shifts equilibrium, with mercury providing stronger buoyancy below the interface.​

Buoyancy Calculation

Initial: Submerged fraction f = ρ_block / 13.6 = 1/4, so ρ_block = 3.4. Final: 3.4 = 13.6x + 1(1-x), solving x = (3.4-1)/(13.6-1) ≈ 0.19. [execute_python]​

CSIR NET Exam Insight

This tests Archimedes’ principle application in layered fluids, common in competitive exams. Options approximate 0.19 to 1/5 (d).​