An adulterated sample of milk has a density `1032 kg m^(-3)` , while pure milk has a density of `1080 kg m^(-3)` . Then the volume of pure milk in a sample of 10 litres of adulterated milk is

Options

A. 1 litre
B. 2 litre
C. 3 litre
D. 4 litre

Explanation

To find the volume of pure milk in a sample of 10 liters of adulterated milk, we can follow these steps:

### Step 1: Understand the problem
We have an adulterated sample of milk with a density of \(1032 \, \text{kg/m}^3\) and pure milk with a density of \(1080 \, \text{kg/m}^3\). We need to find the volume of pure milk in a total volume of \(10 \, \text{liters}\) of the adulterated milk.

### Step 2: Convert the volume of adulterated milk to cubic meters
Since \(1 \, \text{liter} = 10^{-3} \, \text{m}^3\), we can convert \(10 \, \text{liters}\) to cubic meters:
\[ V_A = 10 \, \text{liters} = 10 \times 10^{-3} \, \text{m}^3 = 0.01 \, \text{m}^3 \]

### Step 3: Set up the equation for mass
The mass of the adulterated milk can be calculated using its density:
\[ M_A = \rho_A \times V_A = 1032 \, \text{kg/m}^3 \times 0.01 \, \text{m}^3 = 10.32 \, \text{kg} \]

Let \(V_P\) be the volume of pure milk, and \(V_W\) be the volume of water added. The total volume of the adulterated milk is:
\[ V_A = V_P + V_W \]

### Step 4: Express the mass of pure milk
The mass of pure milk can be expressed as:
\[ M_P = \rho_P \times V_P = 1080 \, \text{kg/m}^3 \times V_P \]

### Step 5: Express the mass of water
The mass of water can be expressed as:
\[ M_W = \rho_W \times V_W = 1000 \, \text{kg/m}^3 \times V_W \]

### Step 6: Relate the masses
Since the mass of the adulterated milk is the sum of the masses of pure milk and water, we have:
\[ M_A = M_P + M_W \]
Substituting the expressions for \(M_A\), \(M_P\), and \(M_W\):
\[ 10.32 = 1080 \times V_P + 1000 \times (0.01 - V_P) \]

### Step 7: Solve for \(V_P\)
Expanding the equation:
\[ 10.32 = 1080 V_P + 10 - 1000 V_P \]
Combining like terms:
\[ 10.32 - 10 = (1080 - 1000) V_P \]
\[ 0.32 = 80 V_P \]
Now, solving for \(V_P\):
\[ V_P = \frac{0.32}{80} = \frac{32}{8000} = \frac{4}{1000} \, \text{m}^3 \]

### Step 8: Convert back to liters
To convert \(V_P\) back to liters:
\[ V_P = \frac{4}{1000} \, \text{m}^3 = 4 \, \text{liters} \]

### Final Answer
The volume of pure milk in the sample of 10 liters of adulterated milk is \(4 \, \text{liters}\).

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