Two wires of same radius having lengths `l_(1)andl_(2)` and resistivities `rho_(1)andrho_(2)` are connected in series. The equivalent resistivity will be

Options

A. `(rho_(1)l_(2)+rho_(2)l_(1))/(rho_(1)+rho_(2))`
B. `(rho_(1)l_(1)+rho_(2)l_(2))/(l_(1)+l_(2))`
C. `(rho_(1)l_(1)+rho_(2)l_(2))/(l_(1)-l_(2))`
D. `(rho_(1)l_(2)+rho_(2)l_(1))/(l_(1)+l_(2))`

Explanation

To find the equivalent resistivity of two wires connected in series, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Setup**: We have two wires of the same radius connected in series. Let the lengths of the wires be \( l_1 \) and \( l_2 \), and their resistivities be \( \rho_1 \) and \( \rho_2 \).

2. **Calculate Resistance of Each Wire**:
The resistance \( R \) of a wire can be calculated using the formula:
\[ R = \frac{\rho L}{A} \]
where \( \rho \) is the resistivity, \( L \) is the length of the wire, and \( A \) is the cross-sectional area. Since both wires have the same radius, their cross-sectional area \( A \) is the same and can be expressed as:
\[ A = \pi R^2 \]
Therefore, the resistances of the two wires can be expressed as:
\[ R_1 = \frac{\rho_1 l_1}{A} \]
\[ R_2 = \frac{\rho_2 l_2}{A} \]

3. **Total Resistance in Series**:
When resistors are connected in series, the total resistance \( R_{eq} \) is the sum of the individual resistances:
\[ R_{eq} = R_1 + R_2 \]
Substituting the expressions for \( R_1 \) and \( R_2 \):
\[ R_{eq} = \frac{\rho_1 l_1}{A} + \frac{\rho_2 l_2}{A} \]

4. **Combine the Terms**:
Since both terms have a common denominator \( A \), we can combine them:
\[ R_{eq} = \frac{\rho_1 l_1 + \rho_2 l_2}{A} \]

5. **Find the Equivalent Resistivity**:
The equivalent resistivity \( \rho_{eq} \) can be defined in terms of the total resistance and the total length of the wire:
\[ R_{eq} = \frac{\rho_{eq} (l_1 + l_2)}{A} \]
Setting the two expressions for \( R_{eq} \) equal gives:
\[ \frac{\rho_{eq} (l_1 + l_2)}{A} = \frac{\rho_1 l_1 + \rho_2 l_2}{A} \]

6. **Cancel the Area**:
Since \( A \) is common in both sides, we can cancel it:
\[ \rho_{eq} (l_1 + l_2) = \rho_1 l_1 + \rho_2 l_2 \]

7. **Solve for Equivalent Resistivity**:
Finally, we can solve for \( \rho_{eq} \):
\[ \rho_{eq} = \frac{\rho_1 l_1 + \rho_2 l_2}{l_1 + l_2} \]

### Final Answer:
The equivalent resistivity of the two wires connected in series is given by:
\[ \rho_{eq} = \frac{\rho_1 l_1 + \rho_2 l_2}{l_1 + l_2} \]