If an increase in length of copper wire is 0.5% due to stretching, the percentage increase in its resistance will be

Options

A. 0.001
B. 0.002
C. 0.01
D. 0.02

Answer

D

Explanation

Given, increase in length of wire, i.e., `(Delta L)/(L)=0.5 %`
We know, resistance `(R )= rho(l)/(A)= rho (l)/(A)xx(A)/(A)= rho (V)/(A^(2)) " "` (`because` volume, i.e., `V=l xx A`)
Percentage increase in resistance `(= rho (V)/(l^(4)))`
(area, i.e., `A = l^(2)`, i.e., l is length of a wire)
i.e., `(Delta R)/(R )xx100=4(Delta l)/(l)xx100=4xx0.5%=2%`