The resistance of a wire of uniform diameter d and length L is R . The resistance of another wire of the same material but diameter `2d` and length 4L will be

Options

A. `2R`
B. `R`
C. `R//2`
D. `R//4`

Answer

To find the resistance of the second wire, we can use the formula for resistance:

\[ R = \frac{\rho L}{A} \]

where:
- \( R \) is the resistance,
- \( \rho \) is the resistivity of the material (constant for the same material),
- \( L \) is the length of the wire,
- \( A \) is the cross-sectional area of the wire.

### Step 1: Identify the parameters for both wires
- For the first wire:
- Length \( L_1 = L \)
- Diameter \( d_1 = d \)
- Cross-sectional area \( A_1 = \pi \left(\frac{d}{2}\right)^2 = \frac{\pi d^2}{4} \)

- For the second wire:
- Length \( L_2 = 4L \)
- Diameter \( d_2 = 2d \)
- Cross-sectional area \( A_2 = \pi \left(\frac{2d}{2}\right)^2 = \pi d^2 \)

### Step 2: Write the resistance formulas for both wires
- Resistance of the first wire:
\[ R_1 = \frac{\rho L_1}{A_1} = \frac{\rho L}{\frac{\pi d^2}{4}} = \frac{4\rho L}{\pi d^2} \]

- Resistance of the second wire:
\[ R_2 = \frac{\rho L_2}{A_2} = \frac{\rho (4L)}{\pi d^2} = \frac{4\rho L}{\pi d^2} \]

### Step 3: Compare the resistances
From the calculations, we see that:
\[ R_1 = \frac{4\rho L}{\pi d^2} \]
\[ R_2 = \frac{4\rho L}{\pi d^2} \]

### Step 4: Conclusion
Since \( R_1 = R_2 \), we conclude that the resistance of the second wire is the same as that of the first wire:
\[ R_2 = R \]

### Final Answer
The resistance of the second wire is \( R \).

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