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

Hint: The volume of the mixture is the sum of the individual volumes and density is the ratio of mass and volume. Use these along with the given information of the mixture. This will help to find equations between the unknown densities. Solve the equations to find densities.

Complete answer:
We have two substances with individual densities as ${\rho _1}$ and ${\rho _2}$.
For the first mixture, we mix the two substances in equal volumes and we get relative density as 4. Let the volume of the first mixture be $V$. Since the individual substances were mixed in equal volumes, individual volumes will be $\dfrac{V}{2}$. The total mass of mixture before and after mixing remains the same and since mass is equal to the product of density and volume.
$
   \Rightarrow {\rho _1}\dfrac{V}{2} + {\rho _2}\dfrac{V}{2} = 4V \\
   \Rightarrow {\rho _1} + {\rho _2} = 8 \\
$
Above equation is the first of relations between the two densities.
For the second mixture, two substances are mixed in equal masses to have a mixture of relative density, 3. Let the individual volumes be ${V_1}$and ${V_2}$. Therefore, ${\rho _1}{V_1} = {\rho _2}{V_2} = \dfrac{{3({V_1} + {V_2})}}{2}$
Consider the first equal pair, we get: $\dfrac{{{\rho _1}}}{{{\rho _2}}} = \dfrac{{{V_2}}}{{{V_1}}}$ and substitute this in the second pair of equal terms.
$
   \Rightarrow {\rho _2} = \dfrac{3}{2}\left( {\dfrac{{{V_1}}}{{{V_2}}} + 1} \right) = \dfrac{3}{2}\left( {\dfrac{{{\rho _2}}}{{{\rho _1}}} + 1} \right) \\
   \Rightarrow 2{\rho _1}{\rho _2} = 3({\rho _1} + {\rho _2}) \\
$
Using the first relation, we get, $ \Rightarrow {\rho _1}{\rho _2} = \dfrac{3}{2} \times 8 = 12 \Rightarrow {\rho _2} = \dfrac{{12}}{{{\rho _1}}}$
From this result and the first relation, we get,
$
   \Rightarrow {\rho _1} + \dfrac{{12}}{{{\rho _1}}} = 8 \\
   \Rightarrow \rho _1^2 - 8{\rho _1} + 12 = 0 \\
   \Rightarrow \rho _1^2 - 6{\rho _1} - 2{\rho _1} - 12 = 0 \\
   \Rightarrow ({\rho _1} - 6)({\rho _1} - 2) = 0 \\
$
We have two values for ${\rho _1}$ and thus for ${\rho _2}$. When ${\rho _1} = 6$, we have ${\rho _2} = 8 - 6 = 2$ and when${\rho _1} = 2$, we have ${\rho _2} = 8 - 2 = 6$. Comparing these results with the given choices, we can say that option A is correct.

Note: Since we have the sum of the densities as constant therefore, we have exchange of values between the densities of the two substances. It is always better to consider the masses will not change before and after the mixture.