If a wire of resistance R is melted and recast into a wire of `(2^(th))/(5)` its original length, then resistance of the new wire will be

Options

A. R
B. `sqrt(2/5)` R
C. `2/5` R
D. `(2/5)^2 R`

Answer

To solve the problem of finding the resistance of a new wire formed by melting and recasting a wire of resistance \( R \) into a wire of \( \frac{2}{5} \) 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.

2. **Identify the change in length**:
The new length \( L' \) of the wire after recasting is:
\[ L' = \frac{2}{5} L \]

3. **Determine the volume of the wire**:
The volume \( V \) of the wire remains constant during the melting and recasting process. The volume of the wire can be expressed as:
\[ V = A \cdot L \]
For the new wire, the volume can also be expressed as:
\[ V = A' \cdot L' \]
where \( A' \) is the new cross-sectional area.

4. **Set the volumes equal**:
Since the volume remains constant, we can equate the two expressions for volume:
\[ A \cdot L = A' \cdot L' \]
Substituting \( L' \):
\[ A \cdot L = A' \cdot \left(\frac{2}{5} L\right) \]
Simplifying this gives:
\[ A' = \frac{5}{2} A \]

5. **Calculate the new resistance**:
Now, we can find the new resistance \( R' \) using the formula for resistance:
\[ R' = \frac{\rho L'}{A'} \]
Substituting for \( L' \) and \( A' \):
\[ R' = \frac{\rho \left(\frac{2}{5} L\right)}{\frac{5}{2} A} \]
Simplifying this:
\[ R' = \frac{\rho \cdot 2L}{5 \cdot \frac{5}{2} A} = \frac{2\rho L}{\frac{25}{2} A} = \frac{4\rho L}{25 A} \]

6. **Relate the new resistance to the original resistance**:
Recall that the original resistance \( R \) is:
\[ R = \frac{\rho L}{A} \]
Therefore, we can express \( R' \) in terms of \( R \):
\[ R' = \frac{4}{25} R \]

### Final Answer:
The resistance of the new wire will be:
\[ R' = \frac{4}{25} R \]