Question

A wire of resistance \(R\) is stretched to double its original length. What is its new resistance?

• A. \(2R\)
• B. \(4R\)
• C. \(\frac{R}{2}\)
• D. \(R\)

Hint:

If a wire is stretched to \(n\) times its original length while volume remains constant, the new resistance becomes \(n^2 R\).

Answer 1

The Correct Option is B

Topic - Current Electricity (Resistance of Conductors):
This question deals with the physical factors affecting the resistance of a conductor, specifically the relationship between length, cross-sectional area, and volume conservation during deformation.
Step 1: Understanding the Question:
We need to determine the new resistance of a wire after it has been stretched to twice its initial length.
Step 2: Key Formula or Approach:
The resistance \(R\) is defined as: \[ R = \rho \frac{L}{A} \] Where:
\(\rho\) = Resistivity of the material.
\(L\) = Length of the conductor.
\(A\) = Area of cross-section.
Since the wire is stretched, its volume \(V = A \times L\) remains constant.
Step 3: Detailed Solution:
1. Let the initial length be \(L\) and area be \(A\).
2. The initial resistance is \(R = \rho \frac{L}{A}\).
3. The new length is \(L' = 2L\).
4. Since volume is constant (\(V = V'\)):
\[ A \cdot L = A' \cdot L' \] \[ A \cdot L = A' \cdot (2L) \] \[ A' = \frac{A}{2} \] 5. The new resistance \(R'\) is:
\[ R' = \rho \frac{L'}{A'} \] \[ R' = \rho \frac{2L}{A/2} \] \[ R' = 4 \left( \rho \frac{L}{A} \right) \] \[ R' = 4R \] Step 4: Final Answer:
The new resistance is \(4R\).