A cylindrical conductor of length `l` and uniform area of cross-section `A` has resistance `R`. Another conductor of length `2 l` and resistance `R` of the same material has area of cross-section :

Options

A. `A//2`
B. `3A//2`
C. `2A`
D. `3A`

Explanation

To solve the problem step by step, we will analyze the relationship between the resistance, length, area of cross-section, and resistivity of the conductors.

### Step 1: Understand the relationship between resistance, length, area, and resistivity
The resistance \( R \) of a cylindrical conductor is given by the formula:
\[ R = \frac{\rho l}{A} \]
where:
- \( R \) is the resistance,
- \( \rho \) is the resistivity of the material,
- \( l \) is the length of the conductor,
- \( A \) is the area of cross-section.

### Step 2: Write the equation for the first conductor
For the first conductor with length \( l \) and area of cross-section \( A \), we have:
\[ R = \frac{\rho l}{A} \tag{1} \]

### Step 3: Write the equation for the second conductor
For the second conductor with length \( 2l \) and resistance \( R \), we denote its area of cross-section as \( A' \). The resistance can be expressed as:
\[ R = \frac{\rho (2l)}{A'} \tag{2} \]

### Step 4: Equate the resistivity from both conductors
Since both conductors are made of the same material, their resistivity \( \rho \) is the same. Therefore, we can equate the two equations (1) and (2):
\[ \frac{\rho l}{A} = \frac{\rho (2l)}{A'} \]

### Step 5: Simplify the equation
We can cancel \( \rho \) and \( l \) from both sides (since they are non-zero):
\[ \frac{1}{A} = \frac{2}{A'} \]

### Step 6: Cross-multiply to find the relationship between areas
Cross-multiplying gives:
\[ A' = 2A \]

### Conclusion
Thus, the area of cross-section \( A' \) of the second conductor is twice that of the first conductor:
\[ A' = 2A \]