The resistivity of aluminium is 2.834 × 10-8 Ω 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)

Question

The resistivity of aluminium is 2.834 × 10-8 Ω 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) 1111S

Tardigrade Admin · Accepted Answer

Radius =(4.0/2) =2.0 mm =2× 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 × ( textLength/ textArea) =(2.834 × 10-8 Ω m× 2.00 m/π(2.0 × 10-3 m)2) = 4.5 × 10-3 Ω Conductance =(1/R) =(1/ 45 × 10-3)=222Ω-1 (S)

Q. The resistivity of aluminium is $2.834 \times 1 0^{- 8} Ω$ 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)

$111.0$ S

$1.11$ S

$222.05$ S

$1111$ S

Q. The resistivity of aluminium is 2.834×10−8Ω 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)

111.0 S

1.11 S

222.05 S

1111S

Radius =24.0​=2.0mm=2×10−3m;l=2m Area of cross section and resistance (R) is related to the resistivity as, (Specific resistance), the length and area by R = Resistivity ×AreaLength​ =π(2.0×10−3m)22.834×10−8Ωm×2.00m​ =4.5×10−3Ω Conductance =R1​=45×10−31​=222Ω−1(S)

Q. The resistivity of aluminium is $2.834 \times 1 0^{- 8} Ω$ 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)

$111.0$ S

$1.11$ S

$222.05$ S

$1111$ S