Question

Two liquids of densities ${\rho }_1$ and ${\rho }_2$ are mixed in two different ways. In first case, equal mass of both the liquids were mixed and the mixture density was found to be $3{\text{kg/m}}^3$ . In second case, equal volume of both the liquids were mixed and the mixture density was found to be $5{\text{kg/m}}^3$ . Then product of densities ${\rho }_1$ and ${\rho }_2$ is

• A. $4{\text{kg}}^2/m^6$
• B. $8{\text{kg}}^2/m^6$
• C. $15{\text{kg}}^2/m^6$
• D. $30{\text{kg}}^2/m^6$

Answer 1

The correct option is C $15{\text{kg}}^2/m^6$

Let $m_1$ and $V_1$ be the mass and volume of fluid of density ${\rho }_1$ and $m_2$ and $V_2$ be the mass and volume of fluid of density ${\rho }_2$
where ${\rho }_1=\frac{m_1}{V_1}$ and ${\rho }_2=\frac{m_2}{V_2}$
If both the fluids are mixed then the density of the mixture will be,
${\rho }_{mix}=\frac{m_1+m_2}{V_1+V_2}$
Case 1: When equal masses of fluids are mixed.
$\begin{array}{}{\rho }_{mix}=\frac{2m}{V_1+V_2}& \text{(1)}\end{array}$
In form of ${\rho }_1$ and ${\rho }_2$, $\left(1\right)$ becomes
${\rho }_{mix1}=\frac{2{\rho }_1{\rho }_2}{{\rho }_1+{\rho }_2}$
Case 2: When equal volume of fluids are mixed.
$\begin{array}{}{\rho }_{mix}=\frac{m_1+m_2}{2V}& \text{(2)}\end{array}$
In form of ${\rho }_1$ and ${\rho }_2$, $\left(2\right)$ becomes
${\rho }_{mix2}=\frac{{\rho }_1+{\rho }_2}{2}$
Now
${\rho }_{mix1}\times {\rho }_{mix2}={\rho }_1\times {\rho }_2$
$\Rightarrow {\rho }_1\times {\rho }_2=3\times 5=15k{g}^2/m^6$