A wire has a resistance of `32Omega`. It is melted and drawn into a wire of half of its original length. What is the resistance of the new wire?

Options

A. `32Omega`
B. `16Omega`
C. `8Omega`
D. `4Omega`

Explanation

To find the resistance of the new wire after it has been melted and drawn into a wire of half its original length, 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:
\[ R = \frac{\rho L}{A} \]
where:
- \( R \) = resistance
- \( \rho \) = resistivity of the material (constant for the same material)
- \( L \) = length of the wire
- \( A \) = cross-sectional area of the wire

### Step 2: Identify the original resistance and length
Given that the original resistance \( R = 32 \, \Omega \), we can express this as:
\[ R = \frac{\rho L}{A} = 32 \, \Omega \]

### Step 3: Determine the new length of the wire
When the wire is melted and drawn into a wire of half its original length, the new length \( L' \) is:
\[ L' = \frac{L}{2} \]

### Step 4: Determine the new area of the wire
Since the volume of the wire remains constant during the melting and drawing process, we can use the relationship between length and area. The volume \( V \) of the wire is given by:
\[ V = L \cdot A \]
For the new wire, the volume must also equal:
\[ V' = L' \cdot A' = \frac{L}{2} \cdot A' \]
Setting the volumes equal gives:
\[ L \cdot A = \frac{L}{2} \cdot A' \]
From this, we can solve for the new area \( A' \):
\[ A' = 2A \]

### Step 5: Calculate the new resistance
Now we can find the new resistance \( R' \) using the new length and area:
\[ R' = \frac{\rho L'}{A'} = \frac{\rho \left(\frac{L}{2}\right)}{2A} \]
This simplifies to:
\[ R' = \frac{\rho L}{4A} \]
Since we know that \( \frac{\rho L}{A} = 32 \, \Omega \), we can substitute this into our equation:
\[ R' = \frac{32}{4} = 8 \, \Omega \]

### Conclusion
The resistance of the new wire is \( 8 \, \Omega \).

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