A copper wire of length 3 m and cross‑sectional area $4\times 10^{-6},{\text{m}}^2$ has a resistivity of $1.68\times 10^{-8},\Omega !\cdot !\text{m}$. Calculate (i) the resistance of the wire at room temperature and (ii) the resistance if the temperature is lowered such that the resistivity decreases by 10%.

Step-by-step solution

1. Use $R=\rho ,L/A$ for the initial wire.
2. Substitute $\rho =1.68\times 10^{-8},\Omega !\cdot !\text{m}$, $L=3,\text{m}$, $A=4\times 10^{-6},{\text{m}}^2$ to get $R=1.68\times 10^{-8}\times 3/4\times 10^{-6}=0.0126,\Omega$.
3. When resistivity decreases by 10 %, new $\rho =0.9\times 1.68\times 10^{-8}=1.512\times 10^{-8},\Omega !\cdot !\text{m}$.
4. Compute new $R=1.512\times 10^{-8}\times 3/4\times 10^{-6}=0.01134,\Omega$.
5. Hence the resistances are 0.0126 Ω and 0.01134 Ω respectively.

Concept tested: Resistivity and Temperature Dependence

Published 7 October 2026 · How this solution was prepared and checked