A wire of resistance `20Omega` is stretched to thrice its original length.
What is the value of new resistance?

Options

A. `120 Omega`
B. `140 Omega`
C. `180 Omega`
D. `100 Omega`

Explanation

To find the new resistance of a wire that has been stretched to thrice 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:
- \( R \) is the resistance,
- \( \rho \) is the resistivity of the material,
- \( l \) is the length of the wire,
- \( A \) is the cross-sectional area of the wire.

### Step 2: Define the original parameters
Let:
- The original length of the wire be \( l \),
- The original cross-sectional area be \( A \),
- The original resistance be \( R = 20 \, \Omega \).

### Step 3: Determine the new length after stretching
When the wire is stretched to thrice its original length, the new length \( l' \) is:
\[ l' = 3l \]

### Step 4: Use the volume conservation principle
Since the volume of the wire remains constant during stretching, we have:
\[ \text{Original Volume} = \text{New Volume} \]
This can be expressed as:
\[ l \cdot A = l' \cdot A' \]
Substituting \( l' = 3l \):
\[ l \cdot A = 3l \cdot A' \]
Cancelling \( l \) from both sides (assuming \( l \neq 0 \)):
\[ A = 3A' \]
Thus, the new cross-sectional area \( A' \) is:
\[ A' = \frac{A}{3} \]

### Step 5: Calculate the new resistance
Now we can find the new resistance \( R' \) using the new length and area:
\[ R' = \frac{\rho l'}{A'} = \frac{\rho (3l)}{\frac{A}{3}} = \frac{3\rho l}{\frac{A}{3}} = 9 \frac{\rho l}{A} \]
Since \( \frac{\rho l}{A} = R \):
\[ R' = 9R \]
Substituting the original resistance \( R = 20 \, \Omega \):
\[ R' = 9 \times 20 = 180 \, \Omega \]

### Conclusion
The new resistance after stretching the wire to thrice its original length is:
\[ \boxed{180 \, \Omega} \]