Question

The electric resistance of a wire is \( R \). If the length of the wire is increased to double by stretching it, then the new resistance of the wire is

• A. \( 2R \)
• B. \( 4R \)
• C. \( R \)
• D. \( 16R \)

Hint:

Stretching wire $\longrightarrow$ length ↑, area ↓ $\longrightarrow$ resistance increases rapidly (here \( \propto L^2 \)).

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Resistance depends on resistivity, length, and cross-sectional area. When a wire is "stretched", its total volume remains constant. Thus, increasing the length necessarily decreases the cross-sectional area.
Step 2: Key Formula or Approach:
1. Resistance: \( R = \rho \frac{L}{A} \).
2. Volume: \( V = A \cdot L = \text{constant} \).
3. Substituting \( A = V/L \) into the resistance formula: \( R = \rho \frac{L^2}{V} \).
4. This implies \( R \propto L^2 \) for a stretched wire.
Step 3: Detailed Explanation:
Initially, resistance is \( R_1 = R \).
The length is doubled, so \( L_2 = 2L_1 \).
Using the proportionality derived in Step 2:
\[ \frac{R_2}{R_1} = \left( \frac{L_2}{L_1} \right)^2 \]
\[ \frac{R_2}{R} = \left( \frac{2L_1}{L_1} \right)^2 = 2^2 = 4 \]
\[ R_2 = 4R \]
The area effectively halved to compensate for the doubling of length, both factors contributing to a four-fold increase in total resistance.
Step 4: Final Answer:
The new resistance of the wire is 4R.