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 solve the problem of finding the volume of pure milk in a 10-liter sample of adulterated milk, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Data:**
- Density of adulterated milk, \( \rho_a = 1032 \, \text{kg/m}^3 \)
- Density of pure milk, \( \rho_p = 1080 \, \text{kg/m}^3 \)
- Volume of adulterated milk, \( V_a = 10 \, \text{liters} = 10 \times 10^{-3} \, \text{m}^3 \)

2. **Calculate the Mass of Adulterated Milk:**
- The mass of the adulterated milk can be calculated using the formula:
\[ m_a = \rho_a \times V_a \]
- Substituting the values:
\[ m_a = 1032 \, \text{kg/m}^3 \times 10 \times 10^{-3} \, \text{m}^3 = 10.32 \, \text{kg} \]

3. **Let \( V_p \) be the Volume of Pure Milk:**
- The mass of pure milk can be expressed as:
\[ m_p = \rho_p \times V_p \]

4. **Set Up the Equation for Mass Balance:**
- The mass of the adulterated milk is equal to the mass of pure milk plus the mass of water added (which we will denote as \( m_w \)):
\[ m_a = m_p + m_w \]
- The mass of water can be expressed as:
\[ m_w = \rho_w \times V_w \]
- Since the density of water \( \rho_w \) is approximately \( 1000 \, \text{kg/m}^3 \) and \( V_w = V_a - V_p \), we can rewrite the equation:
\[ m_a = m_p + \rho_w \times (V_a - V_p) \]

5. **Substituting the Known Values:**
- Substitute \( m_a \) and \( m_p \):
\[ 10.32 = 1080 \times V_p + 1000 \times (10 \times 10^{-3} - V_p) \]

6. **Simplify the Equation:**
- Expanding the equation:
\[ 10.32 = 1080 V_p + 10 - 1000 V_p \]
- Combine like terms:
\[ 10.32 = (1080 - 1000) V_p + 10 \]
\[ 10.32 - 10 = 80 V_p \]
\[ 0.32 = 80 V_p \]

7. **Solve for \( V_p \):**
- Rearranging gives:
\[ V_p = \frac{0.32}{80} = 0.004 \, \text{m}^3 \]

8. **Convert to Liters:**
- To convert cubic meters to liters:
\[ V_p = 0.004 \, \text{m}^3 \times 1000 = 4 \, \text{liters} \]

### Final Answer:
The volume of pure milk in the sample of 10 liters of adulterated milk is **4 liters**.