A wire of resistance `4 Omega` is stretched to twice its original length. In the process of stretching, its area of cross-section gets halved. Now, the resistance of the wire is

Options

A. `8 Omega`
B. `16 Omega`
C. `1 Omega`
D. `4 Omega`

Answer

To find the new resistance of the wire after it has been stretched, we can follow these steps:

### Step 1: Understand the initial conditions
Let the initial resistance of the wire be \( R = 4 \, \Omega \). The initial length of the wire is \( L \) and the initial area of cross-section is \( A \).

### Step 2: Determine the new dimensions after stretching
When the wire is stretched to twice its original length, the new length \( L' \) becomes:
\[ L' = 2L \]
During this process, the area of cross-section is halved, so the new area \( A' \) becomes:
\[ A' = \frac{A}{2} \]

### Step 3: Use 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. The resistivity remains constant during the stretching process.

### Step 4: Calculate the new resistance
The new resistance \( R' \) after stretching can be calculated using the new length and area:
\[ R' = \frac{\rho L'}{A'} \]
Substituting \( L' \) and \( A' \):
\[ R' = \frac{\rho (2L)}{\frac{A}{2}} = \frac{2\rho L}{\frac{A}{2}} = \frac{2\rho L \cdot 2}{A} = \frac{4\rho L}{A} \]

### Step 5: Relate the new resistance to the original resistance
Since the original resistance \( R \) is given by \( R = \frac{\rho L}{A} \), we can substitute this into the equation for \( R' \):
\[ R' = 4 \cdot \frac{\rho L}{A} = 4R \]
Given that \( R = 4 \, \Omega \):
\[ R' = 4 \cdot 4 = 16 \, \Omega \]

### Final Answer
Thus, the new resistance of the wire after stretching is:
\[ \boxed{16 \, \Omega} \]