Question

A metallic sphere floats in an immiscible mixture of water (density 10${}^3$ kg/m${}^3$) and a liquid (density $8\times {10}^{3}kg/m^3$) such that its ($\frac{2}{3}$) part is in water and ($\frac{1}{3}$) part in the liquid. The density of the metal is

• A. $\frac{5000}{3}kg/m^3$
• B. $\frac{10000}{3}kg/m^3$
• C. $5000kg/m^3$
• D. $2000kg/m^3$

Answer 1

The correct option is B

$\frac{10000}{3}kg/m^3$

Obviously water will float on top of the liquid. If the total volume of the ball is V from free body diagram

$mg=F_B+F_{Bq}$

let density of water be ${\rho }_{\omega }$, density of liquid be ${\rho }_l$ and density of ball be $\rho$.

$\rho Vg={\rho }_l\frac{1}{3}Vg+{\rho }_{\omega }\frac{2}{3}Vg$

$\rho =\frac{1}{3}{\rho }_l+\frac{2}{3}{\rho }_{\omega }$

$=\frac{1}{3}\left(8000\right)+\frac{2}{3}\left(1000\right)$

$=\frac{10000}{3}$