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
• E. R/4

Hint:

For stretching problems, use the shortcut $R' = n^2 R$, where $n$ is the factor by which the length is increased. Here, $n=2$, so $2^2 = 4$.

Answer 1

The Correct Option is C

Concept:
Physics - Resistance and Resistivity.
The resistance of a wire is given by $R = \rho \frac{L}{A}$, where $\rho$ is resistivity, $L$ is length, and $A$ is cross-sectional area.
Step 1: Understand the volume conservation.
When a wire is stretched, its volume ($V = A \times L$) remains constant. If the length is doubled ($L' = 2L$), the area must decrease to keep the volume the same: $$ A \times L = A' \times 2L \implies A' = \frac{A}{2} $$
Step 2: Set up the ratio for new resistance.
$$ R_{new} = \rho \frac{L'}{A'} $$
Step 3: Substitute the new dimensions.
$$ R_{new} = \rho \frac{2L}{A/2} $$ $$ R_{new} = 4 \left( \rho \frac{L}{A} \right) $$
Step 4: Conclusion.
$$ R_{new} = 4R $$