Question

Find the density of a metallic body which floats at the interface of mercury of specific gravity $13.6$ and water such that $31.75\%$ of its volume is submerged in mercury and $68.25\%$ in water.

• A. $6000{\text{kg/m}}^3$
• B. $5000{\text{kg/m}}^3$
• C. $4000{\text{kg/m}}^3$
• D. $7000{\text{kg/m}}^3$

Answer 1

The correct option is B $5000{\text{kg/m}}^3$

Given
Specific gravity of mercury, $s_m=13.6$
Volume submerged in the mercury $=31.75\%$
Volume submerged in water $=68.25\%$

Let the volume of the body $=V{\text{m}}^3$
Volume of the body submerged in mercury $=\frac{31.75}{100}V$ $=0.3175V{\text{m}}^3$
and volume of body submerged in water = $\frac{68.25}{100}V$ $=0.6825V{\text{m}}^3$

As we know,
Density of liquid = specific gravity $\times$ ${\rho }_w$
where ${\rho }_w$= density of water
So, density of mercury (${\rho }_m$) $=13.6\times 1000=13600{\text{kg/m}}^3$

Total upward buoyant force ($B$)
= Buoyant force due to water ($B_1$) + Buoyant force due to mercury ($B_2$)
$B_1$ = ${\rho }_w\times g\times 0.6825V$ = $682.5gV$
$B_2$ = ${\rho }_m\times g\times 0.3175V$ = $4318gV$
So, $B$ = $B_1$ + $B_1$ = $5000.5gV$

For the equilibrium of the body,
Total buoyant force (upward force) = Weight of the body
Assume that the density of the body is ${\rho }_b$
$\Rightarrow$$5000.5gV$ = ${\rho }_b\times g\times V$
$\Rightarrow$${\rho }_b$ = $5000.5\text{k}g/m^3$
$\therefore$ Density of body (${\rho }_b)\approx 5000{\text{kg/m}}^3$