Question

what length of nichrome wire of diameter 0.20mm and conductivity 9.1 × 105 ohms-1 m-1, will give a resistance of 10ohms

Answer 1

Given:

Material of wire= Nichrome

Diameter of wire, D= 0.2 mm= 2×10⁻⁴ m

Conductivity of wire, σ= 9.1×10⁵ Ω⁻¹m⁻¹

To Find:

Length of wire such that resistance of wire comes out to be 10Ω

Solution:

We know that,

• $\pink{\boxed{\bf{Radius=\frac{Diameter}{2}}}}$

• Area A of wire having radius r is given by

$\pink{\boxed{\bf{A=\pi r^{2}}}}$

• Resistivity ρ is given by

$\purple{\boxed{\bf{\rho=\frac{1}{\sigma}}}}$

where

σ is conductivity of material

• Resistance R of a conductor is given by

$\purple{\boxed{\bf{R=\frac{\rho l}{A}}}}$

where

ρ is the resistivity of conductor

l is length of conductor

A is area of conductor

$\rule{190}{1}$

Let the radius of wire be 'r'

So,

$\longrightarrow\rm{r=\dfrac{D}{2}}$

$\longrightarrow\rm{r=\dfrac{2\times10^{-4}}{2}}$

$\longrightarrow\rm{r=10^{-4}m}$

$\rule{190}{1}$

Let the Area of given wire be A

So,

$\longrightarrow\rm{A=\pi r^{2}}$

$\longrightarrow\rm{A=\dfrac{22}{7}\times (10^{-4})^{2}}$

$\longrightarrow\rm{A=\dfrac{22}{7}\times 10^{-8}\:m^{2}}$

$\rule{190}{1}$

Let the resistivity of given wire be 'ρ'

So,

$\longrightarrow\rm{\rho=\dfrac{1}{\sigma}}$

$\longrightarrow\rm{\rho=\dfrac{1}{9.1\times10^{5}}}$

$\longrightarrow\rm{\rho=\dfrac{10^{-5}}{9.1}\Omega m}$

$\rule{190}{1}$

Let the length of the given wire be 'l'

So, resistance of given is

$\longrightarrow\rm{R=\dfrac{\rho l}{A}}$

$\longrightarrow\rm{l=\dfrac{RA}{\rho}}$

$\longrightarrow\rm{l=\dfrac{10\times\dfrac{22}{7}\times10^{-8}}{\dfrac{10^{-5}}{9.1}}}$

$\longrightarrow\rm{l=\dfrac{\dfrac{220}{7}\times10^{-8}}{\dfrac{10^{-5}}{9.1}}}$

$\longrightarrow\rm{l=\dfrac{220\times9.1\times10^{-8}}{7\times10^{-5}}}$

$\longrightarrow\rm\green{l=286\times10^{-3}m}$

$\longrightarrow\rm\green{l=0.286\:mm}$

Hence, the required length of given wire should be 286×10⁻³m or 0.286 mm.