Question

A copper wire has diameter 0.5 mm and resistivity of 1.671012 m. What will be
the length of this wire to make its resistance 10 ohm? How much does the resistance
change if the diameter is doubled?

Answer 1

$\huge {\orange{\boxed{ \overline{ \underline{ \mid\mathfrak{\red{Sol}}{\mathrm{\blue{ut}}{ \sf{\green{ion}}} \pink{\colon}\mid}}}}}}$

Given :

• Resistivity (ρ) = 1.671012 Ωm
• Diameter (d) = 0.5 mm = 5*10-³
• Resistance (R) = 10 Ω

____________________________

To Find :

• Length of wire
• Change in resistance if Diameter is doubled.

____________________________

Solution :

• As, Diameter = 0.0005 mm

• Radius 0.00025 m

• And, Area = πr² = 22/7*0.00025 = 1.96 * 10^(-6) m²

We have formula :

$\large \star{\boxed{\sf{R \: = \: \rho \dfrac{l}{A}}}} \\ \\ \implies {\sf{10 \: = \: 1.671012 \: \times \: \dfrac{length}{1.96 \: \times \: 10^{-6}}}} \\ \\ \implies {\sf{length \: = \: \dfrac{10 \: \times \: 1.96 \: \times \: 10^{-6}}{1.671012}}} \\ \\ \implies {\sf{length \: = \: \dfrac{1.96 \: \times \: 10^{-7}}{1.671012}}} \\ \\ \implies {\sf{length \: = \: 1.17 \: \times \: 10^{-7}}}$

Length of wire is 1.17 * 10^(-7) m

$\rule{200}{2}$

If Diameter is doubled then,

• Diameter = 2*0.0005 = 0.001 m
• Radius = 0.0005 m
• length = 1.17*10^(-7) m
• Rho (ρ) = 1.671012 Ωm
• Area = πr² = 22*7 * [1.17*10^(-7)] = 3.67 * 10^(-7)
• R = ?

Use same formula :

$\implies {\sf{R \: = \: \dfrac{1.17 * 10^{-7}\: \times \: 1.671012}{3.67 \: \times \: 10^{-7}}}} \\ \\ \implies {\sf{R \: = \: \dfrac{1.95 \: \times \: 10^{-7}}{3.67 \: \times \: 10^{-7}}}} \\ \\ \implies {\sf{R \: = \: 0.531}} \\ \\ \implies {\boxed{\sf{Resistance \: = \: 0.531 \: \Omega}}}$

Answer 2

Solution :-

Given :-

Resistivity ρ = 1.671012 Ω - m

Diameter of the copper wire d = 0.5 mm

Radius of the copper wire r = d/2 = 0.5/2 = 0.25 mm = 0.00025 m

Area of cross section A = πr² = 3.14 * 0.00025² = 3.14 * 625 * 10^(-10) = 1.9625 * 10^(-7) m²

Length of the wire l = ?

[ Because 1 mm = 1/1000 m ]

Resistance R = 10 Ω

Resistance R = ρl/A

[ Where A = πr²]

Substituting the values in the formula

$\implies \tt 10 = 1.671012 \times \dfrac{l}{1.9625 \times 10^{ - 7} }$

$\implies \tt \dfrac{10 \times 1.9625 \times 10^{ - 7} }{1.671012} =l$

$\implies \tt \dfrac{ 1.9625 \times 10^{ - 6} }{1.671012} =l$

$\implies \tt 1.174 \times 10^{ - 6} =l$

$\implies \tt l = 1.174 \times 10^{ - 6}$

Therefore the length of the wire is 1.174 * 10^(-6) m

If diameter is doubled

Diameter of the copper wire = 2 * 0.0005 m = 0.0001 m

Radius of the copper wire r = 0.0001/2 = 0.0005 m

Area of cross section A = πr² = 3.14 * 0.0005² = 3.14 * 25 * 10^(-8) = 78.5 * 10^(-7) m²

Length of the copper wire l = 1.174 * 10^(-6) m

Resistivity ρ = 1.671012 Ω - m

Resistance R = ρl/A

Substituting the values

$\implies \tt R = 1.671012 \times \dfrac{1.174 \times 10^{ - 6} }{78.5 \times 10^{ - 7} }$

$\implies \tt R = 1.671012 \times \dfrac{1.174 \times 10 }{78.5 }$

$\implies \tt R = 1.671012 \times \dfrac{11.74 }{78.5}$

$\implies \tt R = \dfrac{19.617}{78.5 } = 0.25$

Therefore Resistance when diameter is doubled is 0.25 Ω.