The area of cross-section of rod is given by `A= A_(0) (1+alphax)` where `A_(0)` & `alpha` are constant and x is the distance from one end. If the thermal conductivity of the material is `K`. What is the thermal resistancy of the rod if its length is `l_(0)`?

Options

A. `KA_(0)alphaln(1+alphal_(0))`
B. `(1)/(KA_(0)alpha)ln(1+alphal_(0))`
C. `(alpha)/(KA_(0)alpha)ln(1+alphal_(0))`
D. `(KA_(0))/(alpha)ln(1+alphal_(0))`

Explanation

To find the thermal resistance of the rod with a varying cross-sectional area, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Given Information**:
- The area of cross-section of the rod is given by:
\[ A = A_0 (1 + \alpha x) \]
where \( A_0 \) and \( \alpha \) are constants, and \( x \) is the distance from one end of the rod.
- The thermal conductivity of the material is given as \( K \).
- The length of the rod is \( l_0 \).

2. **Define Thermal Resistance**:
- The thermal resistance \( R \) of a rod can be expressed as:
\[ R = \frac{L}{K \cdot A} \]
where \( L \) is the length of the rod, \( K \) is the thermal conductivity, and \( A \) is the cross-sectional area.

3. **Consider a Differential Element**:
- For a small segment \( dx \) of the rod, the differential thermal resistance \( dR \) can be expressed as:
\[ dR = \frac{dx}{K \cdot A(x)} \]
- Substituting the expression for \( A(x) \):
\[ dR = \frac{dx}{K \cdot A_0 (1 + \alpha x)} \]

4. **Integrate to Find Total Resistance**:
- To find the total resistance \( R \) from \( x = 0 \) to \( x = l_0 \), integrate \( dR \):
\[ R = \int_0^{l_0} \frac{dx}{K \cdot A_0 (1 + \alpha x)} \]
- Since \( K \) and \( A_0 \) are constants, they can be taken out of the integral:
\[ R = \frac{1}{K \cdot A_0} \int_0^{l_0} \frac{dx}{1 + \alpha x} \]

5. **Evaluate the Integral**:
- The integral \( \int \frac{dx}{1 + \alpha x} \) can be solved using the natural logarithm:
\[ \int \frac{dx}{1 + \alpha x} = \frac{1}{\alpha} \ln(1 + \alpha x) \]
- Thus, evaluating from \( 0 \) to \( l_0 \):
\[ R = \frac{1}{K \cdot A_0} \left[ \frac{1}{\alpha} \ln(1 + \alpha l_0) - \frac{1}{\alpha} \ln(1) \right] \]
- Since \( \ln(1) = 0 \):
\[ R = \frac{1}{K \cdot A_0 \cdot \alpha} \ln(1 + \alpha l_0) \]

6. **Final Expression for Thermal Resistance**:
- Therefore, the thermal resistance of the rod is given by:
\[ R = \frac{1}{K \cdot A_0 \cdot \alpha} \ln(1 + \alpha l_0) \]