Question

A copper wire has a diameter of 0.5mm and a resistivity of 1.6*10^-6ohm-cm. How much of this wire would be necessary to make a resistance of 10ohm?

Answer 1

check the attachment ❤️❤️❤️

Answer 2

$\underline{\rm{\green{\huge{Answer:-}}}}$

Length of the wire is 122.77 m

$\rule{100}2$

Explanation:-

A copper wire has a diameter of 0.5 mm and a resistivity of $\sf{1.6\:\times\:10^{-6}}$ ohm cm.

$\underline{\underline{\rm{\red{As\:per\:the\:condition}}}}$

• Diameter of copper wire is 0.5 mm = 5 × 10-⁴ m
• Radius of copper wire is 2.5 × 10-⁴ m
• Resistivity of copper wire is 1.6 × 10^-6 ohm cm = 1.6 × 10^-8 ohm m

We have to find the wire, that how much of the wire is required to make a resistance of 10 ohm.

$\underline{\underline{\rm{\red{As\:per\:the\:condition}}}}$

• Length of the wire =?
• Resistance of wire is 10 ohm

$\underline{\underline{\rm{\red{Solution:-}}}}$

We know that,

$\boxed{ \sf{R \: = \: \rho \times \frac{l}{A}}}$

Substitute the known values above

$\implies\:\sf{10\:=\:1.6\:\times\:10^{-8}\:\dfrac{l}{\pi r^2}}$

$\implies\:\sf{10\:=\:1.6\:\times\:10^{-8}\:\dfrac{l}{\frac{22}{7}\:\times\:(2.5\:\times\:10^{-4})^2}}$

$\implies\:\sf{10\:=\:1.6\:\times\:10^{-8}\:\dfrac{7l}{22\:\times\:(2.5\:\times\:10^{-4})^2}}$

$\implies\:\sf{10\:=\:1.6\:\times\:10^{-8}\:\dfrac{7l}{22\:\times\:6.25\:\times\:10^{-8}}}$

$\implies\:\sf{10\:=\:1.6\:\times\:\dfrac{7l}{22\:\times\:6.25}}$

$\implies\:\sf{ \dfrac{10 \: \times \: 22 \: \times \: 6.25}{1.6 \: \times \: 7\: \: } \: = \: l}$

$\implies\:\sf{\dfrac{1375}{11.2}\:=\:l}$

$\implies\:\sf{122.77\:\:=\:l}$

$\implies\:\sf{l\:=\:122.77\:m}$