Question

Calculate the resistance of a copper wire 1Km long and 0.5mm diameter if the resistivity

of copper is 1.7x10-8Ohm m​

Answer 1

$\color{darkblue}\underline{\underline{\sf Given-}}$

• Lenght ${(\ell)}$ = 1000m
• Resistivity ${\sf (\rho)= 1.7×10^{-8}ohm.m}$
• Diameter (d) = 0.5m

$\underline{\underline{\sf Radius \: of \: wire-}}$

$\implies{\sf r = \dfrac{d}{2} }$

$\implies{\sf r = \dfrac{0.5}{2}}$

$\implies{\sf r = \dfrac{1}{4}mm}$

$\implies{\sf r = 0.25mm }$

$\implies{\sf \orange{r = 0.25×10^{-3}m }}$

$\color{darkblue}\underline{\underline{\sf To \: Find-}}$

• Resistance of copper wire (R).

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

$\color{green}\underline{\underline{\sf Formula \: Used-}}$

$\Large\color{violet}\blacksquare\underline{\boxed{\sf R = \rho \dfrac{\ell}{A}}}$

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

$\implies{\sf R = 1.7×10^{-8}×\dfrac{1000}{πr^2} }$

$\implies{\sf R = 1.7×10^{-8}×\dfrac{1000}{3.14×(0.25×10^{-3})^2}}$

$\implies{\sf R =1.7×10^{-8}×\dfrac{1000}{3.14×625×10^{-4}×10^{-6}}}$

$\implies{\sf R = 1.7×10^{-8}×\dfrac{1000}{1962.5×10^{-10}} }$

$\implies{\sf R = \dfrac{1700×10^2}{1962.5}}$

$\color{red}\implies{\sf R = 86.6m }$

$\color{darkblue}\underline{\underline{\sf Answer-}}$

Resistance of copper wire is $\color{red}{\sf R = 86.6m}$.

Answer 2

Explanation:

Given−

Lenght {(\ell)}(ℓ) = 1000m

Resistivity {\sf (\rho)= 1.7×10^{-8}ohm.m}(ρ)=1.7×10

−8

ohm.m

Diameter (d) = 0.5m

\underline{\underline{\sf Radius \: of \: wire-}}

Radiusofwire−

\implies{\sf r = \dfrac{d}{2} }⟹r=

2

d

\implies{\sf r = \dfrac{0.5}{2}}⟹r=

2

0.5

\implies{\sf r = \dfrac{1}{4}mm}⟹r=

4

1

mm

\implies{\sf r = 0.25mm }⟹r=0.25mm

\implies{\sf \orange{r = 0.25×10^{-3}m }}⟹r=0.25×10

−3

m

\color{darkblue}\underline{\underline{\sf To \: Find-}}

ToFind−

Resistance of copper wire (R).

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

\color{green}\underline{\underline{\sf Formula \:. Used-}}

FormulaUsed−

\Large\color{violet}\blacksquare\underline{\boxed{\sf R = \rho \dfrac{\ell}{A}}}■

R=ρ

A

ℓ

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

\implies{\sf R = 1.7×10^{-8}×\dfrac{1000}{πr^2} }⟹R=1.7×10

−8

×

πr

2

1000

\implies{\sf R = 1.7×10^{-8}×\dfrac{1000}{3.14×(0.25×10^{-3})^2}}⟹R=1.7×10

−8

×

3.14×(0.25×10

−3

)

2

1000

\implies{\sf R =1.7×10^{-8}×\dfrac{1000}{3.14×625×10^{-4}×10^{-6}}}⟹R=1.7×10

−8

×

3.14×625×10

−4

×10

−6

1000

\implies{\sf R = 1.7×10^{-8}×\dfrac{1000}{1962.5×10^{-10}} }⟹R=1.7×10

−8

×

1962.5×10

−10

1000

\implies{\sf R = \dfrac{1700×10^2}{1962.5}}⟹R=

1962.5

1700×10

2

\color{red}\implies{\sf R = 86.6m }⟹R=86.6m

\color{darkblue}\underline{\underline{\sf Answer-}}

Answer−

Resistance of copper wire is \color{red}{\sf R = 86.6m}R=86.6m .