A copper wire of length l and radius r is nickel plated till its final radius is 2r. If the resistivity of the copper and nickel are `rho_(c)` and `rho_(n)`, then find the equivalent resistance of the wire.

Explanation

To find the equivalent resistance of a copper wire that has been nickel plated, we can follow these steps:

### Step 1: Calculate the Resistance of the Copper Wire
The resistance \( R_c \) of the copper wire can be calculated using the formula:
\[ R_c = \frac{\rho_c \cdot l}{A_c} \]
where:
- \( \rho_c \) is the resistivity of copper,
- \( l \) is the length of the wire,
- \( A_c \) is the cross-sectional area of the copper wire.

The cross-sectional area \( A_c \) of the copper wire with radius \( r \) is given by:
\[ A_c = \pi r^2 \]
Thus, the resistance of the copper wire becomes:
\[ R_c = \frac{\rho_c \cdot l}{\pi r^2} \]

### Step 2: Calculate the Resistance of the Nickel Plated Section
The nickel plating increases the radius to \( 2r \). The resistance \( R_n \) of the nickel section can be calculated similarly:
\[ R_n = \frac{\rho_n \cdot l}{A_n} \]
where:
- \( \rho_n \) is the resistivity of nickel,
- \( A_n \) is the cross-sectional area of the nickel plated wire.

The cross-sectional area \( A_n \) of the nickel section with radius \( 2r \) is:
\[ A_n = \pi (2r)^2 = \pi \cdot 4r^2 \]
Thus, the resistance of the nickel section becomes:
\[ R_n = \frac{\rho_n \cdot l}{\pi \cdot 4r^2} \]

### Step 3: Combine the Resistances in Parallel
Since the copper wire and the nickel plated section are in parallel, the equivalent resistance \( R_{eq} \) can be calculated using the formula for resistances in parallel:
\[ \frac{1}{R_{eq}} = \frac{1}{R_c} + \frac{1}{R_n} \]
Substituting the expressions for \( R_c \) and \( R_n \):
\[ \frac{1}{R_{eq}} = \frac{\pi r^2}{\rho_c \cdot l} + \frac{\pi \cdot 4r^2}{\rho_n \cdot l} \]
Factoring out common terms:
\[ \frac{1}{R_{eq}} = \frac{1}{l} \left( \frac{\pi r^2}{\rho_c} + \frac{4\pi r^2}{\rho_n} \right) \]
\[ \frac{1}{R_{eq}} = \frac{\pi r^2}{l} \left( \frac{1}{\rho_c} + \frac{4}{\rho_n} \right) \]

### Step 4: Solve for \( R_{eq} \)
Taking the reciprocal to find \( R_{eq} \):
\[ R_{eq} = \frac{l}{\pi r^2} \cdot \frac{1}{\left( \frac{1}{\rho_c} + \frac{4}{\rho_n} \right)} \]
This can be simplified further:
\[ R_{eq} = \frac{l \cdot \rho_c \cdot \rho_n}{\pi r^2 \left( \rho_n + 4\rho_c \right)} \]

### Final Result
The equivalent resistance of the wire is:
\[ R_{eq} = \frac{\rho_c \cdot \rho_n \cdot l}{\pi r^2 \left( \rho_n + 4\rho_c \right)} \]