Question

A cylinder wire 0.5m in length and 0.5mm in diameter, has a resistance of 2.5 ohms. Find the resistivity of the wire

Answer 1

$\large \rm {\underbrace{\underline{Elucidation:-}}}$

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$\sf \red {\underline{\underline{Provided\: that:-}}}$

➻Length of the wire(l)=0.5m

➻Diamter of the wire(d)=0.5mm

➻d=0.0005m

➻Resistance (R)=2.5Ω

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$\sf \blue {\underline{\underline{To\: Determine:-}}}$

➻Resistivity of the wire=?

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$\sf \pink {\underline{\underline{Resistivity:-}}}$

➻Definition:- Resistivity also called "specific resistance" of a material is the resistance of a wire of that material of unit length and unit area of cross section.It is denoted by "ρ".

➻Units:- ohm×metre(Ωm)

➻Dimensional formula:- $\tt {M^{1}\: L^{3}\: T^{-3}\: I^{-2}}$

➻By definition,

➻The formula for calculating resistivity ,

$\longrightarrow \tt {\boxed{\underline{ρ=\frac{RA}{l}}}}$

➻where,

➻R denotes resistance

➻A denotes Area of cross section

➻L denotes length.

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➻To calculate resistivity,we need to find area of cross section.

$\sf \purple {\underline{\underline{We\: know:-}}}$

➻Area of cross section of a cylinder=$\tt {πr^{2}}$

➻We've been given the diameter.

➻Using "d=2r",

$\to \tt {diameter=2\times radius}$

$\to \tt {0.0005m=2\times radius}$

$\to \tt {radius=\frac{0.0005}{2}}$

$\to \tt \green {\fbox{radius(r)=0.00025m}}$

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➻As Radius is determined we can calculate area of cross section(A).

$\to \tt {Area\: of\: cross\: section(A)=πr^{2}}$

$\to \tt {A=\frac{22}{7}\times (0.00025)^{2}}$

$\to \tt {A=\frac{22}{7}\times 0.00025 \times 0.00025}$

$\to \tt \green {\fbox{A≈1.964}}$

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➻So , finally

$\large \mapsto \tt {ρ=\frac{RA}{l}}$

➻Supplanting the given values,

$\large \mapsto \tt {ρ=\frac{2.5Ω\times1.96m}{0.5}}$

$\large \mapsto \tt {ρ=\frac{\cancel{2.5}^{5}Ω\times1.96m}{\cancel{0.5}}}$

$\mapsto \tt {ρ=5\times 1.96}$

$\large \mapsto \tt {ρ=\frac{5\times 196}{100}}$

$\large \mapsto \tt {ρ=\frac{\cancel{5}\times 196}{\cancel{100}_{20}}}$

$\large \mapsto \tt {ρ=\frac{196}{20}}$

$\large \mapsto \tt {ρ=\frac{\cancel{196}^{98}}{\cancel{20}_{10}}}$

$\large \mapsto \tt {ρ=\frac{98}{10}}$

$\implies \tt \green {\fbox{ρ=9.8Ωm}}$

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$\sf \orange {\underline{\underline{Thereupon,}}}$

➻The resistivity of the wire is "9.8Ωm" respectively.

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