Question

A wire of resistance 2R is stretched such that its length is doubled. Then the increase in its resistance is: 

• A. 6R
• B. 4R
• C. 3R
• D. 2R

Answer 1

The Correct Option is A

To solve the problem, we need to determine how the resistance of a wire changes when its length is doubled, while its volume remains constant.

1. Understanding the Resistance Formula:
Resistance of a wire is given by:

$R = \rho \frac{L}{A}$
Where:
- $R$ is the resistance,
- $\rho$ is the resistivity (constant for the material),
- $L$ is the length,
- $A$ is the cross-sectional area.

2. Initial Resistance:
We are given that the initial resistance of the wire is $2R$.

3. When the Wire is Stretched:
Let the original length be $L$ and area be $A$. When stretched to double its length ($2L$), and assuming volume remains constant:
Volume before = Volume after:
$A \cdot L = A' \cdot 2L$
$\Rightarrow A' = \frac{A}{2}$

4. New Resistance:
New resistance $R_{\text{new}}$ is:
$R_{\text{new}} = \rho \cdot \frac{2L}{A/2} = \rho \cdot \frac{4L}{A}$

We know original resistance $R_0 = \rho \cdot \frac{L}{A}$. So:
$R_{\text{new}} = 4 \cdot R_0 = 4 \cdot (2R) = 8R$

5. Increase in Resistance:
Increase = $R_{\text{new}} - R_0 = 8R - 2R = 6R$

Final Answer:
The increase in resistance is $6R$.