Question

If length of a conductor is doubled by keeping volume constant, then what is its new resistance if initial were 4 Ω?

• A. 16 Ω
• B. 8 Ω
• C. 4Ω
• D. 2Ω

Answer 1

Answer: (A)

16 Ω

Answer 2

A conductor of length $L$ and cross‐sectional area $A$ has resistance

$R=\rho L/A$.

If its length is doubled to $2L$ while the volume $V=AL$ remains constant, then

$A'=\frac{AL}{2L}=\frac{A}{2}$.

Thus the new resistance is

$R'=\rho (2L)/(A/2)=4\rho L/A=4R$.

Since the original resistance is 4 Ω,

$R' =16Ω$.