If a wire of resistance R is melted and recasted to half of its length, then the new resistance of the wire will be

Options

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

Answer

To solve the problem of finding the new resistance of a wire that is melted and recast to half of its length, we can follow these steps:

### Step 1: Understand the initial conditions
Let the initial length of the wire be \( L \) and its cross-sectional area be \( A \). The initial resistance \( R \) of the wire can be expressed using the formula:
\[ R = \frac{\rho L}{A} \]
where \( \rho \) is the resistivity of the material.

### Step 2: Determine the new length and area
When the wire is melted and recast to half of its length, the new length \( L' \) becomes:
\[ L' = \frac{L}{2} \]
Since the volume of the wire remains constant during the melting and recasting process, we can use the volume conservation principle. The initial volume \( V \) of the wire is:
\[ V = L \times A \]
The new volume \( V' \) after recasting is:
\[ V' = L' \times A' = \frac{L}{2} \times A' \]
Setting the initial and new volumes equal gives:
\[ L \times A = \frac{L}{2} \times A' \]

### Step 3: Solve for the new cross-sectional area
From the volume conservation equation, we can solve for the new cross-sectional area \( A' \):
\[ A' = 2A \]

### Step 4: Calculate the new resistance
Now we can calculate the new resistance \( R' \) using the formula for resistance:
\[ R' = \frac{\rho L'}{A'} = \frac{\rho \left(\frac{L}{2}\right)}{2A} \]
Substituting \( L' \) and \( A' \) into the resistance formula gives:
\[ R' = \frac{\rho \frac{L}{2}}{2A} = \frac{\rho L}{4A} \]
Since \( R = \frac{\rho L}{A} \), we can express \( R' \) in terms of \( R \):
\[ R' = \frac{R}{4} \]

### Conclusion
Thus, the new resistance of the wire after it has been melted and recast to half of its length is:
\[ \boxed{\frac{R}{4}} \]