Let us consider a copper wire having radius r and length l. Let its resistance be R. If the radius of another copper wire is `2r` and the length is `l//2` then the resistance of this, wire will be

Options

A. R
B. 2R
C. `R//4`
D. `R//8`

Explanation

To find the resistance of the second copper wire, we will use the formula for the resistance of a cylindrical conductor:

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

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

### Step 1: Calculate the cross-sectional area of the first wire
The cross-sectional area \( A_1 \) of the first wire with radius \( r \) is given by:

\[ A_1 = \pi r^2 \]

### Step 2: Calculate the resistance of the first wire
Using the formula for resistance, the resistance \( R \) of the first wire can be expressed as:

\[ R = \rho \frac{l}{A_1} = \rho \frac{l}{\pi r^2} \]

### Step 3: Calculate the cross-sectional area of the second wire
The radius of the second wire is \( 2r \). Therefore, the cross-sectional area \( A_2 \) of the second wire is:

\[ A_2 = \pi (2r)^2 = \pi (4r^2) = 4\pi r^2 \]

### Step 4: Calculate the resistance of the second wire
The length of the second wire is \( \frac{l}{2} \). Using the resistance formula, the resistance \( R_2 \) of the second wire is:

\[ R_2 = \rho \frac{\frac{l}{2}}{A_2} = \rho \frac{\frac{l}{2}}{4\pi r^2} \]

### Step 5: Simplify the expression for \( R_2 \)
Now, substituting \( A_2 \):

\[ R_2 = \rho \frac{l}{2 \cdot 4\pi r^2} = \rho \frac{l}{8\pi r^2} \]

### Step 6: Relate \( R_2 \) to \( R \)
From the expression for \( R \):

\[ R = \rho \frac{l}{\pi r^2} \]

Now, we can express \( R_2 \) in terms of \( R \):

\[ R_2 = \frac{1}{8} \left( \rho \frac{l}{\pi r^2} \right) = \frac{1}{8} R \]

### Final Answer
Thus, the resistance of the second wire is:

\[ R_2 = \frac{R}{8} \]