The resistance of a wire is `5Omega`.It's new resistance in ohm if stretched to `5` times of its original length will be

Options

A. `125`
B. `5`
C. `25`
D. `625`

Answer

To find the new resistance of a wire when it is stretched to five times its original length, we can use the relationship between resistance, length, and cross-sectional area.

### Step-by-Step Solution:

1. **Understanding the Relationship**:
The resistance \( R \) of a wire is given by the formula:
\[ R = \rho \frac{L}{A} \]
where \( R \) is the resistance, \( \rho \) is the resistivity of the material, \( L \) is the length of the wire, and \( A \) is the cross-sectional area.

2. **Initial Conditions**:
Let the original length of the wire be \( L_1 \) and the original cross-sectional area be \( A_1 \). The original resistance is given as:
\[ R_1 = 5 \, \Omega \]

3. **New Length After Stretching**:
When the wire is stretched to five times its original length, the new length \( L_2 \) becomes:
\[ L_2 = 5L_1 \]

4. **Volume Conservation**:
The volume of the wire remains constant during stretching. Therefore, we have:
\[ L_1 A_1 = L_2 A_2 \]
Substituting \( L_2 \):
\[ L_1 A_1 = (5L_1) A_2 \]
Dividing both sides by \( L_1 \):
\[ A_1 = 5 A_2 \implies A_2 = \frac{A_1}{5} \]

5. **New Resistance Calculation**:
Now we can find the new resistance \( R_2 \) using the new length and area:
\[ R_2 = \rho \frac{L_2}{A_2} = \rho \frac{5L_1}{\frac{A_1}{5}} = \rho \frac{5L_1 \cdot 5}{A_1} = 25 \left(\rho \frac{L_1}{A_1}\right) = 25 R_1 \]
Since \( R_1 = 5 \, \Omega \):
\[ R_2 = 25 \times 5 = 125 \, \Omega \]

6. **Final Answer**:
Therefore, the new resistance of the wire after stretching it to five times its original length is:
\[ R_2 = 125 \, \Omega \]