A wire of resistance `20Omega` is stretched to thrice its original length.
What is the change in the resistivity of the wire?

Options

A. doubled
B. remains same
C. halved
D. quadrupled

Explanation

To solve the problem, we need to understand the relationship between resistance, resistivity, length, and area of a wire. 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,
- \( L \) is the length of the wire,
- \( A \) is the cross-sectional area of the wire.

### Step 1: Understand the given data
We are given:
- Initial resistance \( R = 20 \, \Omega \)
- The wire is stretched to three times its original length, so if the original length is \( L \), the new length \( L' = 3L \).

### Step 2: Analyze the effect of stretching on dimensions
When a wire is stretched to three times its original length, its volume remains constant. The volume \( V \) of the wire can be expressed as:

\[ V = A \cdot L \]

After stretching, the new volume \( V' \) is:

\[ V' = A' \cdot L' = A' \cdot (3L) \]

Since the volume remains constant, we can equate the two volumes:

\[ A \cdot L = A' \cdot (3L) \]

### Step 3: Solve for the new area
From the equation above, we can simplify to find the new area \( A' \):

\[ A' = \frac{A}{3} \]

### Step 4: Calculate the new resistance
Now, we can find the new resistance \( R' \) using the new length and area:

\[ R' = \frac{\rho L'}{A'} = \frac{\rho (3L)}{(A/3)} = \frac{3\rho L \cdot 3}{A} = 9 \cdot \frac{\rho L}{A} = 9R \]

Since the original resistance \( R = 20 \, \Omega \), the new resistance \( R' \) becomes:

\[ R' = 9 \cdot 20 = 180 \, \Omega \]

### Step 5: Determine the change in resistivity
However, resistivity \( \rho \) is a property of the material and does not change with the dimensions of the wire. Therefore, despite the changes in length and area, the resistivity remains constant.

### Conclusion
The change in resistivity of the wire is:

\[ \Delta \rho = 0 \]