Question

A cubical iron block of side 10 cm is floating on mercury in a vessel.What is the height of the block above the mercury level?
Given that density of iron is 7.2 gm/c.c. and density of mercury = 13.6 gm/c.c.

Answer 1

Given,

Density of iron, ${\rho }_i=7.2gm/cc$

Density of mercury, ${\rho }_m=13.6gm/cc$

Volume of iron cube, $V_c=10\times 10\times 10={10}^{3}c{m}^3$

Area of one face of cube, $A=10\times 10={10}^{2}c{m}^2$

Height of cube, $H=10cm$

Let

Height of cube above mercury, $X$

Area of one face of cube, $A$

For floating condition, Archimedes’ principle

Buoyancy force = weight of mercury displace by cube = weight of iron cube

${\rho }_{m}A(10-X)={\rho }_c{V}_c$

$X=10-\frac{{\rho }_c{V}_c}{{\rho }_{m}A}=10-\frac{7.2\times {10}^3}{13.6\times {10}^2}$

$X=4.7cm$