The temperature coefficient of resistance of conductor varies as `alpha(T) = 3T^2 +2T.` If `R_0` is resistance at T = 0and R is resistance at T, then

Options

A. `R = R_0(6T - 2)`
B. `R = 2R_0(3 +2T0`
C. `R = R_0(1 + T^2 +T^3)`
D. `R = R-0(1 - T + T^2 T^3)`

Explanation

To solve the problem, we need to find the resistance \( R \) at temperature \( T \) given the temperature coefficient of resistance \( \alpha(T) = 3T^2 + 2T \) and the resistance \( R_0 \) at \( T = 0 \).

### Step-by-Step Solution:

1. **Understanding the relationship between resistance and temperature coefficient**:
The temperature coefficient of resistance \( \alpha \) is defined as:
\[ \alpha = \frac{dR}{R_0 \, dT} \]
Rearranging gives:
\[ dR = \alpha R_0 \, dT \]

2. **Substituting the given expression for \( \alpha \)**:
Substitute \( \alpha(T) = 3T^2 + 2T \) into the equation:
\[ dR = (3T^2 + 2T) R_0 \, dT \]

3. **Integrating both sides**:
We need to integrate \( dR \) from \( R_0 \) to \( R \) and \( dT \) from \( 0 \) to \( T \):
\[ \int_{R_0}^{R} dR = \int_{0}^{T} (3T^2 + 2T) R_0 \, dT \]

4. **Calculating the integral**:
The left side becomes:
\[ R - R_0 \]
For the right side, we can factor out \( R_0 \):
\[ R_0 \int_{0}^{T} (3T^2 + 2T) \, dT \]
Now, calculate the integral:
\[ \int (3T^2 + 2T) \, dT = \left[ T^3 + T^2 \right]_{0}^{T} = T^3 + T^2 \]
Therefore, the right side becomes:
\[ R_0 (T^3 + T^2) \]

5. **Equating both sides**:
Now we have:
\[ R - R_0 = R_0 (T^3 + T^2) \]
Rearranging gives:
\[ R = R_0 + R_0 (T^3 + T^2) \]

6. **Factoring out \( R_0 \)**:
\[ R = R_0 (1 + T^3 + T^2) \]

### Final Result:
Thus, the expression for resistance \( R \) at temperature \( T \) is:
\[ R = R_0 (1 + T^2 + T^3) \]