If a copper wire is stretched to make it 0.1% longer, What is the percentage change in its resistance?
A) 0.2%
B) 0.5%
C) 0.1%
D) 0.05%

Hint: A uniform cylinder of resistance L is made of a material with resistance R, cross-sectional area A, and resistivity $\rho$ is given by R = \rho \times \dfrac{l}{A}$.

Complete step by step solution:
To find the change in resistance let’s assume , Length of wire = L
Area of cross- section = A and Resistance = R
As given wire is stretched by 0.1% so after increase
New length of wire = L’
New are of cross section = A’
New resistance = R’
$\eqalign{
  & {\text{L' = L + 0}}{\text{.1% of L}} \cr
  & \Rightarrow {\text{L' = L + }}\dfrac{{0.1}}{{100}} \times {\text{L}} \cr
  & \Rightarrow {\text{L' = 1}}{\text{.001L}} \cr
  & {\text{But, the volume remains same,}} \cr
  & \Rightarrow {\text{V' = V}} \cr
  & \Rightarrow {\text{A'L' = AL}} \cr
  & \Rightarrow {\text{A'(1001)L = AL}} \cr
  & \Rightarrow {\text{A' = }}\dfrac{{\text{A}}}{{1.001}} \cr
  & \therefore {\text{ New resistance R' = }}\rho \times \dfrac{{L'}}{{A'}} \cr
  & \Rightarrow {\text{R' = }}\rho {\text{ }} \times {\text{ }}\dfrac{{1.001{\text{ }} \times {\text{ L}}}}{{\dfrac{{\text{A}}}{{1.001}}}} \cr
  & \Rightarrow {\text{R' = (1}}{\text{.001}}{{\text{)}}^2}.{\text{R}} \cr
  & {\text{Now percentage change in resistance }} \cr
  & \dfrac{{{\text{R' - R}}}}{{\text{R}}}{\text{ }} \times {\text{ 100}} \cr
  & \Rightarrow \dfrac{{{\text{(1}}{\text{.002)R - R}}}}{{\text{R}}}{\text{ }} \times {\text{ 100% }} \cr
  & \Rightarrow {\text{0}}{\text{.002 }} \times {\text{ 100% }} \cr
  & \Rightarrow {\text{0}}{\text{.2% }} \cr
  & \therefore {\text{Resistance will be increased by 0}}{\text{.2% }} \cr} $.

Hence option A is correct.

Notes: The resistance of a transmitter is directly proportional to its length (L, So doubling its length will double its resistance, while making its length half it will halve its resistance.