Question

The resistivity of aluminium is $2.834 \times 10^{-8} \Omega$ m. Thus, conductance across a piece of aluminium wire, that is $4.0$ mm in diameter and $2.00$ m long is (assume current $= 1.25$ A)

• A. $111.0$ S
• B. $1.11$ S
• C. $222.05$ S
• D. $1111$S

Answer 1

Answer: (C)

Radius $=\frac{4.0}{2} =2.0\,\,mm\,\,=2\times 10^{-3} m;\,\, l=2\,m$
Area of cross section and resistance (R) is related to the resistivity as,
(Specific resistance), the length and area by
R = Resistivity $ \times \frac{\text{Length}}{\text{Area}}$
$=\frac{2.834 \times 10^{-8} \Omega m\times 2.00 \,m}{\pi(2.0 \times 10^{-3} m)^{2}}$
$= 4.5 \times 10^{-3} \Omega $
Conductance $=\frac{1}{R} =\frac{1}{ 45 \times 10^{-3}}=222\Omega^{-1} (S)$