Question

A metallic sphere floats in an immiscible mixture of water and a liquid such that its $\frac{4}{5}\text{th}$ volume is in water and $\frac{1}{5}\text{th}$ volume is in the liquid. Then, density of the metal is

• A. $3.5\times {10}^3{\text{kg/m}}^3$
• B. $1.5\times {10}^3{\text{kg/m}}^3$
• C. $4\times {10}^3{\text{kg/m}}^3$
• D. $2\times {10}^3{\text{kg/m}}^3$

Answer 1

The correct option is A $3.5\times {10}^3{\text{kg/m}}^3$

From Archimede's principle,
$\text{Total Buoyant force}=\text{Weight of displaced fluid}$
This buoyant force will balance weight of the sphere
$\therefore \text{Weight of the ball}=\text{Weight of displaced fluid}$
Buoyant force due to immersion in both liquids will balance the weight of sphere.
${\rho }_{metal}gV={\rho }_{w}g\times \left(\frac{4}{5}V\right)+{\rho }_{L}g\times \left(\frac{1}{5}V\right)$
[For sphere, $W=mg={\rho }_{metal}gV$]
$\Rightarrow {\rho }_{metal}=\frac{4}{5}{\rho }_w+\frac{1}{5}{\rho }_L$
Substituting the given data,
$\Rightarrow {\rho }_{metal}=\frac{4}{5}\times {10}^3+\frac{1}{5}\times 13.5\times {10}^3$
$\therefore {\rho }_{metal}=3.5\times {10}^3{\text{kg/m}}^3$