Question

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

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

Answer 1

The correct option is A

${\rho }_1=6;{\rho }_2=2$

Relative density

The ratio of the density of a substance to the density of a standard is called relative density.

Step 1: Calculation of the relative density of the mixture when mixed in equal volume

$$\begin{gathered} \frac{{\rho }_1+{\rho }_2}{2}=4 \\ {\rho }_1+{\rho }_2=8......\left(1\right) \end{gathered}$$

Where, ${\rho }_1$ is the density of the first substance and ${\rho }_2$ is the density of the second substance.

Step 2: Calculation of the relative density of the mixture when mixed in equal mass

$$\begin{gathered} \frac{2{\rho }_1{\rho }_2}{{\rho }_1+{\rho }_2}=3 \\ \frac{2{\rho }_1{\rho }_2}{8}=3 \\ {\rho }_1{\rho }_2=12 \\ {\rho }_1=\frac{12}{{\rho }_2} \end{gathered}$$

Step 3: Calculation of densities for both substance

$$\begin{gathered} \frac{12}{{\rho }_2}+{\rho }_2=8 \\ \frac{12+{\rho }_2^2}{{\rho }_2}=8 \\ {\rho }_2^2-8{\rho }_2+12=0 \\ {\rho }_2=2 \\ Fromequation\left(1\right), \\ {\rho }_1=8-2=6 \end{gathered}$$

Hence, the densities of the substances are ${\rho }_1=6;{\rho }_2=2$.

Therefore, option (A) is correct.