`A` wire of resistance `160Omega` is melted and drawn in a wire of one fourth of its length. The new resistance of the wire will be

Options

A. `40Omega`
B. `16Omega`
C. `10Omega`
D. `640Omega`

Answer

To find the new resistance of a wire after it has been melted and drawn to one-fourth of its original length, we can follow these steps:

### Step-by-Step Solution:

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 \( \rho \) is the resistivity of the material, \( L \) is the length of the wire, and \( A \) is the cross-sectional area of the wire.

2. **Identify the original resistance and length**:
We are given that the original resistance \( R = 160 \, \Omega \). Let the original length of the wire be \( L \) and the original cross-sectional area be \( A \).

3. **Determine the new length**:
The wire is melted and drawn to one-fourth of its original length. Therefore, the new length \( L' \) is:
\[ L' = \frac{L}{4} \]

4. **Calculate the volume of the wire**:
The volume of the wire remains constant when it is melted and reshaped. The volume \( V \) of the wire can be expressed as:
\[ V = L \cdot A \]
After reshaping, the new volume \( V' \) is:
\[ V' = L' \cdot A' = \frac{L}{4} \cdot A' \]
Since the volume remains constant, we have:
\[ L \cdot A = \frac{L}{4} \cdot A' \]

5. **Solve for the new area**:
Rearranging the volume equation gives:
\[ A' = 4A \]
This means the new cross-sectional area is four times the original area.

6. **Substitute into the resistance formula**:
Now, we can find the new resistance \( R' \):
\[ R' = \frac{\rho L'}{A'} = \frac{\rho \left(\frac{L}{4}\right)}{4A} = \frac{\rho L}{16A} \]

7. **Relate the new resistance to the original resistance**:
Since the original resistance \( R = \frac{\rho L}{A} = 160 \, \Omega \), we can substitute this into our equation for \( R' \):
\[ R' = \frac{R}{16} = \frac{160}{16} = 10 \, \Omega \]

### Final Answer:
The new resistance of the wire will be \( 10 \, \Omega \).

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