An electric wire of length ‘ I ’ and area of cross-section a has a resistance R ohm s. Another wire of the same material having same length and area of cross-section 4a has a resistance of

Options

A. `4R`
B. `R//4`
C. `R//16`
D. `16R`

Answer

To solve the problem, we need to determine the resistance of the second wire, which has the same length and material as the first wire but a different cross-sectional area.

### Step-by-Step Solution:

1. **Understand the Formula for Resistance**:
The resistance \( R \) of a wire is given by the formula:
\[ R = \frac{\rho L}{A} \]
where \( \rho \) is the resistivity of the material, \( L \) is the length of the wire, and \( A \) is the cross-sectional area.

2. **Identify Given Values**:
- For the first wire:
- Length \( L = I \)
- Area of cross-section \( A = a \)
- Resistance \( R_1 = R \)
- For the second wire:
- Length \( L = I \) (same as the first wire)
- Area of cross-section \( A = 4a \)

3. **Calculate Resistance of the Second Wire**:
Using the resistance formula for the second wire:
\[ R_2 = \frac{\rho L}{A} \]
Substituting the values for the second wire:
\[ R_2 = \frac{\rho I}{4a} \]

4. **Relate the Resistances**:
Since the resistivity \( \rho \) and length \( L \) are the same for both wires, we can express the resistance of the second wire in terms of the first wire's resistance:
\[ R_1 = \frac{\rho I}{a} \]
Therefore, we can express \( \rho I \) as:
\[ \rho I = R_1 \cdot a \]
Now substituting this into the equation for \( R_2 \):
\[ R_2 = \frac{R_1 \cdot a}{4a} = \frac{R_1}{4} \]

5. **Substituting Known Values**:
Since \( R_1 = R \):
\[ R_2 = \frac{R}{4} \]

### Final Answer:
The resistance of the second wire is:
\[ R_2 = \frac{R}{4} \text{ ohms} \]