Q. A certain electrical conductor has a square cross-section, 2.0 mm on side, and is 12 m long. The resistance between its ends is $0.072\Omega$. The resistivity of its material is equal to
A) $2.4 \times 10^{- 6}\Omega m$
B) $1.2 \times 10^{- 6}\Omega m$
C) $1.2 \times 10^{- 8}\Omega m$
D) $2.4 \times 10^{- 8}\Omega m$
Explanation
(d) The resistance R of a particular conductor is related to the resistivity $\rho$ of its material by
$R = \frac{\rho l}{A}$
Or
$\rho = resistivity\ = \frac{RA}{l}$
Given, R=0.072$\Omega$
$A = 2mm \times 2mm = 4 \times 10^{- 6}m^{2},\ l$=12
$\therefore\ \ \ \ \rho = \frac{0.072 \times 4 \times 10^{- 6}}{12}$
$= 2.4 \times 10^{- 8}\Omega m$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.