Question

A wire of resistance $R$ is stretched to double its length. Its new resistance will be:

• A. $R$
• B. $2R$
• C. $4R$
• D. $R/2$

Hint:

For stretching: $R \propto l^{2}$. If length is $n$ times, resistance is $n^{2}$ times.

Answer 1

The Correct Option is C

Step 1: Concept
Resistance $R = \rho l / A$. When stretched, volume $V = Al$ remains constant.
Step 2: Analysis
If $l' = 2l$, then $A' = A/2$ to keep volume constant. $R' = \rho (2l) / (A/2) = 4 (\rho l / A)$.
Step 3: Conclusion
The new resistance is $4R$.
Final Answer: (C)