Question

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

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

Hint:

Remember that resistance is directly proportional to the length of the conductor and inversely proportional to the cross-sectional area.

Answer 1

The Correct Option is A

Step 1: Recall the resistance formula.
The resistance of a wire is \[ R = \rho\frac{L}{A} \] where $\rho$ is the resistivity, $L$ is the length, and $A$ is the cross-section area.

Step 2: Note what stays fixed.
When a wire is stretched, the material is the same, so $\rho$ does not change. Also no material is lost, so the volume $V = AL$ stays the same.

Step 3: Track the length and area.
The length doubles, so it becomes $2L$. Since the volume is fixed, the area must halve to keep $A\times L$ the same. So the new area is $A/2$.

Step 4: Put in the new values.
\[ R' = \rho\frac{2L}{A/2} = \rho\frac{2L\cdot 2}{A} = 4\rho\frac{L}{A} \]

Step 5: Compare with the start.
Since $R = \rho\dfrac{L}{A}$, we get \[ R' = 4R \]

Step 6: Remember the rule.
When a wire is stretched $n$ times longer, the resistance grows $n^2$ times. Here $n=2$, so it grows $4$ times. \[ \boxed{R' = 4R} \]