A metallic wire of resistance 20 ohm stretched until its length is doubled. Its resistances is

Options

A. `20Omega`
B. `40Omega`
C. `80Omega`
D. `60Omega`

Explanation

To solve the problem of finding the resistance of a metallic wire after it has been stretched to double its length, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the initial resistance formula**:
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 initial conditions**:
We are given that the initial resistance \( R \) is 20 ohms. Therefore, we can write:
\[ R = 20 \, \Omega \]

3. **Consider the effect of stretching the wire**:
When the wire is stretched to double its length, the new length \( L' \) becomes:
\[ L' = 2L \]

4. **Volume conservation**:
The volume of the wire before and after stretching remains constant. Thus, we have:
\[ A \cdot L = A' \cdot L' \]
Substituting \( L' = 2L \):
\[ A \cdot L = A' \cdot (2L) \]
This simplifies to:
\[ A' = \frac{A}{2} \]
So, the new cross-sectional area \( A' \) is half of the original area.

5. **Calculate the new resistance**:
Now we can find the new resistance \( R' \) using the new length and area:
\[ R' = \frac{\rho L'}{A'} \]
Substituting \( L' = 2L \) and \( A' = \frac{A}{2} \):
\[ R' = \frac{\rho (2L)}{\frac{A}{2}} = \frac{2\rho L \cdot 2}{A} = \frac{4\rho L}{A} \]
From the initial resistance formula, we know \( \frac{\rho L}{A} = R \):
\[ R' = 4R \]

6. **Substituting the initial resistance**:
Since \( R = 20 \, \Omega \):
\[ R' = 4 \cdot 20 = 80 \, \Omega \]

### Final Answer:
The new resistance after stretching the wire is \( 80 \, \Omega \).

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