An Aluminium `(alpha=4xx10^(-3)K^(-1))` resistance `R_(1)` and a carbon `(alpha=-0.5xx10^(-3)K^(-1))` resistance `R_(2)` are connected in series to have a resultant resistance of `36Omega` at all temperatures. The values of `R_(1)` and `R_(2)` in `Omega` respectively are:

Options

A. 32, 4
B. 16, 20
C. 4, 32
D. 20, 16

Explanation

To solve the problem, we need to find the values of the resistances \( R_1 \) (Aluminium) and \( R_2 \) (Carbon) given that they are connected in series and their total resistance is \( 36 \, \Omega \) at all temperatures. We also know the temperature coefficients of resistance for both materials.

### Step-by-Step Solution:

1. **Understand the Given Information:**
- Temperature coefficient of Aluminium, \( \alpha_1 = 4 \times 10^{-3} \, \text{K}^{-1} \)
- Temperature coefficient of Carbon, \( \alpha_2 = -0.5 \times 10^{-3} \, \text{K}^{-1} \)
- Total resistance in series, \( R_1 + R_2 = 36 \, \Omega \)

2. **Write the Equation for Change in Resistance:**
- The change in resistance for each material can be expressed as:
\[ \Delta R_1 = R_1 \alpha_1 \Delta T \]
\[ \Delta R_2 = R_2 \alpha_2 \Delta T \]
- Since the total resistance is constant at all temperatures, the overall change in resistance must be zero:
\[ \Delta R_1 + \Delta R_2 = 0 \]

3. **Set Up the Equation:**
- Substituting the expressions for \( \Delta R_1 \) and \( \Delta R_2 \) into the equation:
\[ R_1 \alpha_1 \Delta T + R_2 \alpha_2 \Delta T = 0 \]
- Factoring out \( \Delta T \) (assuming \( \Delta T \neq 0 \)):
\[ R_1 \alpha_1 + R_2 \alpha_2 = 0 \]

4. **Substitute the Values of \( \alpha_1 \) and \( \alpha_2 \):**
- Plugging in the values:
\[ R_1 (4 \times 10^{-3}) + R_2 (-0.5 \times 10^{-3}) = 0 \]
- This simplifies to:
\[ 4 R_1 - 0.5 R_2 = 0 \]
- Rearranging gives:
\[ 4 R_1 = 0.5 R_2 \quad \Rightarrow \quad R_2 = 8 R_1 \]

5. **Substitute \( R_2 \) into the Total Resistance Equation:**
- Using the total resistance equation \( R_1 + R_2 = 36 \):
\[ R_1 + 8 R_1 = 36 \]
- This simplifies to:
\[ 9 R_1 = 36 \]
- Solving for \( R_1 \):
\[ R_1 = \frac{36}{9} = 4 \, \Omega \]

6. **Find \( R_2 \):**
- Now substituting \( R_1 \) back to find \( R_2 \):
\[ R_2 = 8 R_1 = 8 \times 4 = 32 \, \Omega \]

### Final Answer:
The values of \( R_1 \) and \( R_2 \) are:
\[ R_1 = 4 \, \Omega, \quad R_2 = 32 \, \Omega \]