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 values of `rho_(1)` and `rho_(2)` are:

Options

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

Explanation

To solve the problem, we need to find the densities \( \rho_1 \) and \( \rho_2 \) of two substances based on the information given about their mixtures in equal volumes and equal masses.

### Step 1: Define the Variables
Let:
- \( \rho_1 \) = density of substance 1
- \( \rho_2 \) = density of substance 2

### Step 2: Mixture in Equal Volume
When mixed in equal volumes, the relative density of the mixture is given as 4.

The formula for relative density (specific gravity) is:
\[ \text{Relative Density} = \frac{\text{Density of Mixture}}{\text{Density of Water}} \]

Assuming the density of water is \( 1 \, \text{g/cm}^3 \), we can express the density of the mixture when mixed in equal volumes as:
\[ \rho_m = \frac{\rho_1 + \rho_2}{2} \]
Given that the relative density of the mixture is 4:
\[ \frac{\rho_1 + \rho_2}{2} = 4 \implies \rho_1 + \rho_2 = 8 \quad (1) \]

### Step 3: Mixture in Equal Masses
When mixed in equal masses, the relative density of the mixture is given as 3.

Let’s denote the mass of each substance as \( m \). The volume of each substance when mixed in equal masses is:
\[ V_1 = \frac{m}{\rho_1}, \quad V_2 = \frac{m}{\rho_2} \]
The total volume of the mixture is:
\[ V_m = V_1 + V_2 = \frac{m}{\rho_1} + \frac{m}{\rho_2} \]
The density of the mixture when mixed in equal masses is:
\[ \rho_m = \frac{2m}{V_m} = \frac{2m}{\frac{m}{\rho_1} + \frac{m}{\rho_2}} = \frac{2}{\frac{1}{\rho_1} + \frac{1}{\rho_2}} \]
Given that the relative density of the mixture is 3:
\[ \frac{2}{\frac{1}{\rho_1} + \frac{1}{\rho_2}} = 3 \implies \frac{1}{\rho_1} + \frac{1}{\rho_2} = \frac{2}{3} \quad (2) \]

### Step 4: Solve the Equations
We now have two equations:
1. \( \rho_1 + \rho_2 = 8 \)
2. \( \frac{1}{\rho_1} + \frac{1}{\rho_2} = \frac{2}{3} \)

From equation (2), we can express it as:
\[ \frac{\rho_1 + \rho_2}{\rho_1 \rho_2} = \frac{2}{3} \]
Substituting \( \rho_1 + \rho_2 = 8 \) into this equation gives:
\[ \frac{8}{\rho_1 \rho_2} = \frac{2}{3} \implies \rho_1 \rho_2 = 12 \quad (3) \]

### Step 5: Solve for \( \rho_1 \) and \( \rho_2 \)
Now we have two equations (1) and (3):
1. \( \rho_1 + \rho_2 = 8 \)
2. \( \rho_1 \rho_2 = 12 \)

Let \( \rho_1 \) and \( \rho_2 \) be the roots of the quadratic equation:
\[ x^2 - (8)x + 12 = 0 \]
Using the quadratic formula:
\[ x = \frac{8 \pm \sqrt{8^2 - 4 \cdot 12}}{2} = \frac{8 \pm \sqrt{64 - 48}}{2} = \frac{8 \pm \sqrt{16}}{2} = \frac{8 \pm 4}{2} \]
Thus, we have:
\[ x = \frac{12}{2} = 6 \quad \text{or} \quad x = \frac{4}{2} = 2 \]
So, \( \rho_1 = 6 \) and \( \rho_2 = 2 \) (or vice versa).

### Final Answer
The values of \( \rho_1 \) and \( \rho_2 \) are:
\[ \rho_1 = 6 \, \text{g/cm}^3, \quad \rho_2 = 2 \, \text{g/cm}^3 \]