If a wire of resistance `R` is melted and recasted in 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, we need to determine the new resistance of a wire after it has been melted and recast into half of its original length. Let's go through the steps systematically.

### Step 1: Understand the initial conditions
We start with a wire of resistance \( R \), length \( L \), and cross-sectional area \( A \). The resistance of the wire is given by the formula:

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

where \( \rho \) is the resistivity of the material.

### Step 2: Determine the volume of the wire
The volume \( V \) of the wire can be expressed as:

\[ V = A \times L \]

### Step 3: Melting and recasting the wire
When the wire is melted and recast into half of its original length, the new length \( L' \) becomes:

\[ L' = \frac{L}{2} \]

Since the volume remains constant during this process, we can express the new area of cross-section \( A' \) in terms of the original area \( A \):

\[ V = A \times L = A' \times L' \]

Substituting \( L' \):

\[ A \times L = A' \times \frac{L}{2} \]

### Step 4: Solve for the new area of cross-section
Rearranging the equation gives:

\[ A' = \frac{2A}{1} = 2A \]

### Step 5: Calculate the new resistance
Now we can find the new resistance \( R' \) using the resistance formula with the new length and area:

\[ R' = \frac{\rho L'}{A'} = \frac{\rho \left(\frac{L}{2}\right)}{2A} \]

### Step 6: Simplify the expression
Substituting \( R = \frac{\rho L}{A} \) into the equation for \( R' \):

\[ R' = \frac{\frac{1}{2} \rho L}{2A} = \frac{\rho L}{4A} = \frac{R}{4} \]

### Conclusion
Thus, the new resistance \( R' \) of the wire after it has been melted and recast into half of its length is:

\[ R' = \frac{R}{4} \]

### Final Answer
The new resistance of the wire will be \( \frac{R}{4} \).

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