Question

Two substances of relative densities ${\rho }_1$ and ${\rho }_2$ are mixed in equal volumes and the relative density of the mixture is $4$. When they are mixed in equal masses, the relative density of the mixture is $3$. The values of ${\rho }_1$ and ${\rho }_2$ are

• A. ${\rho }_1=6$ and ${\rho }_2=2$
• B. ${\rho }_1=3$ and ${\rho }_2=5$
• C. ${\rho }_1=12$ and ${\rho }_2=4$
• D. None of these

Answer 1

The correct option is A ${\rho }_1=6$ and ${\rho }_2=2$

As we know,
$\text{Density}=\frac{\text{Total Mass}}{\text{Total volume}}=\frac{M}{V}$

and $\text{Mass}=\text{Volume}\times \text{Density}$

When both substances are mixed in equal volume,
$\rho =\frac{m_1+m_2}{v_1+v_2}$
$\Rightarrow 4=\frac{{\rho }_{1}V+{\rho }_{2}V}{2V}$
$\Rightarrow {\rho }_1+{\rho }_2=8$ ... (i)

When both substances are mixed in equal masses:-
$\rho =\frac{m_1+m_2}{v_1+v_2}$
$\Rightarrow 3=\frac{m+m}{\left(m/{\rho }_1\right)+\left(m/{\rho }_2\right)}$
$\Rightarrow \frac{{\rho }_1{\rho }_2}{{\rho }_1+{\rho }_2}=\frac{3}{2}$ ... (ii)
Solving these two equations, we get ${\rho }_1=6$ and ${\rho }_2=2$