Question

Calculate the area of cross section of a wire of length 2m, its resistance is 25Ω and the resistivity of material of wire is 1.84*10-6 Ωm​

Answer 1

Answer :-

Area of cross section of the wire is 1.472 × 10⁻⁷ m² .

Explanation :-

We have :-

→ Length of the wire = 2 m

→ Resistance = 25 Ω

→ Resistivity = 1.84 × 10⁻⁶ Ωm

________________________________

We know that :-

R = ρL/A

Where :-

• R is the resistance.

• ρ is resistivity.

• L is length of the wire.

• A is area of cross section of the wire.

Substituting values, we get :-

⇒ 25 = [1.84 × 10⁻⁶ × 2]/A

⇒ 25A = 3.68 × 10⁻⁶

⇒ A = [3.68 × 10⁻⁶]/25

⇒ A = 0.1472 × 10⁻⁶

⇒ A = 1.472 × 10⁻⁷ m²

Answer 2

Given : Length ( l ) of wire is 2 meters , it's Resistance ( R ) is 25 Ω & resistivity ( $\rho$ ) of material of wire is 1.84 × 10$^{-6}$ Ω.m

Exigency To Find : Area ( A ) of Cross section of wire .

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀

⠀⠀⠀⠀⠀▪︎⠀⠀Finding AREA of CROSS SECTION of wire :

⠀Given that ,

⠀⠀⠀⠀⠀▪︎⠀⠀Length ( L ) of wire is 2 meters ,

⠀⠀⠀⠀⠀▪︎⠀⠀it's Resistance ( R ) is 25 Ω &

⠀⠀⠀⠀⠀▪︎⠀⠀Resistivity( $\bf \rho$ ) of material of wire is 1.84*10$^{-6}$ Ω.m .

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

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

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

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

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

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

$\qquad:\implies \sf 25 \:\:= \: \dfrac{1.84 \: \times 10^{-6}\times 2 }{A} \:\:$

$\qquad:\implies \sf 25 \:\:= \: \dfrac{3.68 \: \times 10^{-6} }{A} \:\:$

$\qquad:\implies \sf 25A \:\:= \: 3.68 \: \times 10^{-6} \:\:$

$\qquad:\implies \sf A \:\:= \:\dfrac{ 3.68 \: \times 10^{-6} }{25} \:\:$

$\qquad:\implies \sf A \:\:= \:\dfrac{ \cancel {3.68} \: \times 10^{-6} }{\cancel {25}} \:\:$

$\qquad:\implies \sf A \:\:= \:0.1472 \: \times 10^{-6} \:\:$

$\qquad:\implies \sf A \:\:= \:1.472 \: \times 10^{-7} \:\:$

$\qquad:\implies \bf A \:\:= \:1.472 \: \times 10^{-7} \: m^2 \:\:$

$\qquad :\implies \pmb{\underline{\purple{\:A \:\:= \:1.472 \: \times 10^{-7} \: m^2 }} }\:\bigstar$

⠀⠀⠀⠀⠀▪︎⠀⠀⠀Here , A denotes Area of Cross section of wire which is $\bf \:1.472 \: \times 10^{-7} \: m^2$

Therefore,

⠀⠀⠀⠀⠀$\therefore {\underline{ \mathrm {\:Hence,\:Area \:of\:Cross \:Section \:of\:wire\:is\:\bf{\:1.472 \: \times 10^{-7} \: m^2 }}.}}$