Question

Resistance of a metal wire of length 1 m is 26Ω at 20∘C. If the diameter of the wire is 0.2 mm, what will be the resistivity of the metal at that temperature?​

Answer 1

Answer :-

Resistivity of the metal wire at that temperature is 8.164 × 10⁻⁷ Ωm .

Explanation :-

We have :-

→ Length of the wire = 1 m

→ Resistance = 26 Ω

→ Diameter of the wire = 0.2 mm = 2 × 10⁻⁴ m

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Firstly, let's calculate the radius of the wire :-

= 2 × 10⁻⁴/2

= 10⁻⁴ m

Area of cross section of the wire :-

⇒ A = πr²

⇒ A = 3.14 × 10⁻⁴ × 10⁻⁴

⇒ A = 3.14 × 10⁻⁸ m²

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Now, we know that :-

R = ρl/A

Substituting values, we get :-

⇒ 26 = (ρ × 1)/3.14 × 10⁻⁸

⇒ ρ = 26 × 3.14 × 10⁻⁸

⇒ ρ = 81.64 × 10⁻⁸

⇒ ρ = 8.164 × 10⁻⁷ Ωm

Answer 2

Given : Resistance of a metal wire of length 1 m is 26Ω at 20⁰C & the diameter of the wire is 0.2 mm .

Exigency To Find : Resistivity of metal at that temperature ?

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀

⠀⠀⠀⠀⠀Given that ,

⠀⠀⠀⠀⠀⠀⠀▪︎⠀⠀Length of metal wire , L = 1 m

⠀⠀⠀⠀⠀⠀⠀▪︎⠀ Resistance of metal wire is , R = 26 Ω

⠀⠀⠀⠀⠀⠀⠀▪︎⠀⠀Diameter of the wire is 0.2 mm or 2 × $\bf 10^{-4}$ m

Now ,

⠀⠀⠀⠀⠀⠀⠀▪︎⠀RADIUS of wire :

$\qquad:\implies \sf \:Radius _{(Wire)} \:=\:\dfrac{Diameter}{2}$

$\qquad:\implies\sf \:Radius _{(Wire)} \:=\:\dfrac{2 \times 10^{-4} }{2}$

$\qquad:\implies\sf \:Radius _{(Wire)} \:=\:\dfrac{\cancel {2} \times 10^{-4} }{\cancel {2}}$

$\qquad:\implies\sf \:Radius _{(Wire)} \:=\: 10^{-4}$

$\qquad \therefore \pmb{\underline{\purple{\:\frak{ \:Radius _{(Wire)} \:=\: 10^{-4} \:m }}} }\:\:\bigstar$

Now ,

⠀⠀⠀⠀⠀⠀⠀▪︎⠀AREA of Cross Section of wire :

$\qquad:\implies \sf \:Area_{(Cros \:Section \:of \: Wire)} \:=\:\pi \times ( Radius )^2$

$\qquad:\implies \sf \:Area_{(Cros \:Section \:of \: Wire)} \:=\:\pi \times \bigg( 10^{-4} \bigg)^2$

$\qquad:\implies \sf \:Area_{(Cros \:Section \:of \: Wire)} \:=\:\pi \times 10^{-8}$

$\qquad:\implies \sf \:Area_{(Cros \:Section \:of \: Wire)} \:=\:3.14 \times 10^{-8}$

$\qquad \therefore \pmb{\underline{\purple{\:\frak{ Area_{(Cros \:Section \:of \: Wire)} \:=\:3.14 \times 10^{-8}\:m^2 }}} }\:\:\bigstar$

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀

$\dag\:\:\pmb{ As,\:We\:know\:that\::}\\\\ \qquad\maltese\bf \:\: Resistance\:of\:wire \:\::$

$\qquad \dag\:\:\bigg\lgroup \sf{ \qquad R \:=\: \dfrac{\rho \:\times L }{A } \qquad }\bigg\rgroup$

⠀⠀⠀⠀⠀ Here , $\rho$ is the Resistivity of metal , R is the Resistance of wire , L is the Length of wire & A is the Area of cross section of wire.

$\qquad \dashrightarrow \sf R \:=\: \dfrac{\rho \:\times L }{A }$

⠀⠀⠀⠀⠀⠀$\underline {\boldsymbol{\star\:Now \: By \: Substituting \: the \: known \: Values \::}}$

$\qquad \dashrightarrow \sf 26 \:=\: \dfrac{\rho \:\times 1 }{ 3.14 \times 10^{-8} }$

$\qquad \dashrightarrow \sf 26 \:=\: \dfrac{\rho \:}{ 3.14 \times 10^{-8} }$

$\qquad \dashrightarrow \sf \: \rho \:=\:3.14\times 26 \times 10^{-8}$

$\qquad \dashrightarrow \sf \: \rho \:=\:81.64 \times 10^{-8}$

$\qquad \dashrightarrow \sf \: \rho \:=\:8.164 \times 10^{-7}$

$\qquad \therefore \pmb{\underline{\purple{\:\frak{ \rho \:=\:8.164 \times 10^{-7}\:\Omega . m }}} }\:\:\bigstar$

$\therefore \:\underline {\sf Hence, \:The \:resistivity \:of \: metal \: , \:\rho \:=\: \bf \:8.164 \times 10^{-7}\:\Omega . m}$