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

Options

A. 16R
B. 2R
C. 4R
D. 0

Explanation

To find the resistance of a wire after it is stretched such that its radius becomes \( \frac{r}{2} \), we can follow these steps:

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

### Step 2: Define the initial conditions
Let:
- Initial radius of the wire \( r_1 = r \)
- Initial length of the wire \( L_1 = L \)
- Initial area \( A_1 = \pi r_1^2 = \pi r^2 \)

### Step 3: Define the final conditions after stretching
After stretching, the new radius is:
\[ r_2 = \frac{r}{2} \]
The new area can be calculated as:
\[ A_2 = \pi r_2^2 = \pi \left(\frac{r}{2}\right)^2 = \pi \frac{r^2}{4} \]

### Step 4: Use the volume conservation principle
The volume of the wire remains constant before and after stretching. Therefore:
\[ L_1 A_1 = L_2 A_2 \]
Substituting the areas:
\[ L \cdot \pi r^2 = L_2 \cdot \pi \frac{r^2}{4} \]
Cancelling \( \pi r^2 \) from both sides gives:
\[ L = \frac{L_2}{4} \]
This implies:
\[ L_2 = 4L \]

### Step 5: Calculate the initial and final resistances
The initial resistance \( R_1 \) is:
\[ R_1 = \rho \frac{L}{A_1} = \rho \frac{L}{\pi r^2} \]
The final resistance \( R_2 \) after stretching is:
\[ R_2 = \rho \frac{L_2}{A_2} = \rho \frac{4L}{\pi \frac{r^2}{4}} = \rho \frac{4L \cdot 4}{\pi r^2} = \rho \frac{16L}{\pi r^2} \]

### Step 6: Relate the final resistance to the initial resistance
Now, we can express \( R_2 \) in terms of \( R_1 \):
\[ R_2 = 16 \cdot R_1 \]
This means the final resistance \( R_2 \) is 16 times the initial resistance \( R_1 \).

### Conclusion
Thus, when the wire is stretched and its radius becomes \( \frac{r}{2} \), its resistance will be:
\[ R_2 = 16 R_1 \]