Resistance of a metal wire of length $1 m$ is $26 Ω$ at $20^{\circ}$ C. If the diameter of the wire is $0.3 m m$ , what will be the resistivity of the metal at that temperature?

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Solution

$Given:$ $R = 26 Ω$ , $l = 1 m$ , $d = 0.3 m m$

$Step 1: Resistivity of the metal (ρ)$
We know that resistance can be given as :
$R = \frac{ρ l}{A}$
$\Rightarrow ρ = \frac{R A}{l}$ $. . . . (1)$

$Step 2: Calculations and Substituting the values$

$A = \frac{π d^{2}}{4} = 3.14 \times \frac{(0.3 \times 10^{- 3} m)^{2}}{4} = 0.071 \times 10^{- 6} m^{2}$
From equation (1)
$ρ = \frac{26 Ω \times 0.071 \times 10^{- 6} m^{2}}{1 m} = 1.85 \times 10^{- 6} Ω m$

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Similar questions

(a) Define resistivity. Write an expression for the resistivity of a substance. Give the meaning of each symbol which occurs in it.
(b) State the SI unit of resistivity.
(c) Distinguish between resistance and resistivity.
(d) Name two factors on which the resistivity of a substance depends and two factors on which it does not depend.
(e) The resistance of a metal wire for length $1 m$ is $26 o h m$ at $2 0^{o}$ C, If the diameter of the wire is $0.3 mm$ , what will be the resistivity of the metal at that temperature?

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Resistance of a metal wire of length 1 m is 26Ω at 20∘C. If the diameter of the wire is 0.3 mm, what will be the resistivity of the metal at that temperature?

Given: R=26 Ω, l=1m, d=0.3mmStep 1: Resistivity of the metal (ρ)We know that resistance can be given as :R=ρ lA⇒ρ=R Al ....(1)Step 2: Calculations and Substituting the values A=πd24=3.14×(0.3×10−3m)24=0.071×10−6m2From equation (1) ρ=26 Ω×0.071×10−6 m21 m=1.85×10−6Ωm

Resistance of a metal wire of length $1 m$ is $26 Ω$ at $20^{\circ}$ C. If the diameter of the wire is $0.3 m m$ , what will be the resistivity of the metal at that temperature?