When a wire is stretched and its radius becomes `r//2` then its resistance will be

Options

A. zero
B. `2R`
C. `8 R`
D. `16 R`

Explanation

To solve the problem of how the resistance of a wire changes when its radius is reduced to half, we can follow these steps:

### Step 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,
- \( A \) is the cross-sectional area of the wire.

### Step 2: Determine the initial cross-sectional area
The cross-sectional area \( A \) of a wire with radius \( r \) is given by:
\[ A = \pi r^2 \]

### Step 3: Calculate the initial resistance
Substituting the expression for area into the resistance formula, we have:
\[ R = \frac{\rho L}{\pi r^2} \]

### Step 4: Analyze the change in radius
When the radius is reduced to \( \frac{r}{2} \), the new cross-sectional area \( A' \) becomes:
\[ A' = \pi \left(\frac{r}{2}\right)^2 = \pi \frac{r^2}{4} \]

### Step 5: Calculate the new resistance
The resistance with the new radius can be calculated as:
\[ R' = \frac{\rho L'}{A'} \]
where \( L' \) is the new length of the wire after stretching. Since the volume of the wire remains constant, we have:
\[ A \cdot L = A' \cdot L' \]
Substituting the areas:
\[ \pi r^2 \cdot L = \pi \frac{r^2}{4} \cdot L' \]
This simplifies to:
\[ L' = 4L \]

### Step 6: Substitute back to find new resistance
Now substituting \( L' \) and \( A' \) back into the resistance formula:
\[ R' = \frac{\rho (4L)}{\pi \frac{r^2}{4}} = \frac{4\rho L}{\frac{\pi r^2}{4}} = \frac{16 \rho L}{\pi r^2} \]

### Step 7: Relate new resistance to original resistance
From the original resistance \( R = \frac{\rho L}{\pi r^2} \), we can express the new resistance as:
\[ R' = 16R \]

### Conclusion
Thus, when the radius of the wire is reduced to half, the resistance becomes:
\[ R' = 16R \]