Question

a wire has diameter of 0.7mm and resistivity of 1.6×10^8 ohm meter. what will be the length of this wire to make its resistance 14 ohm​

Answer 1

Given :

a wire has diameter of 0.7 mm and resistivity of 1.6×10⁻⁸ Ω meter

To Find :

Length of the wire to make its resistance 14 Ω

Solution :

$\bigstar\ \; \sf \pink{R=\dfrac{\rho\ l}{A}}$

where ,

• R denotes resistance
• ρ denotes resistivity
• l denotes length
• A denotes area

_____________________

Given ,

Diameter = 0.7 mm

⇒ Radius , r = 0.35 × 10⁻³ m

So , Area of the wire is ( Circular ) ,

⇒ A = πr²

⇒ A = π ( 0.35 × 10⁻³ )²

⇒ A = 0.1225π × 10⁻⁶ m²

Resistivity , ρ = 1.6 × 10⁻⁸ Ω-m

Resistance , R = 14 Ω

$\\ \to \sf R=\dfrac{\rho\ l}{A}$

$\\ \to \sf 14=\dfrac{1.6 \times 10^{-8} \times l}{0.1225\times \pi \times 10^{-6}}$

$\\ \to \sf 14=\dfrac{1.6\times 10^{-2}\times l}{0.1225\times \pi}$

$\\ \to \sf 5.3851=1.6\times 10^{-2}\times l$

$\\ \to \sf 3.365 = 10^{-2}\times l$

$\\ \to \sf 336.5 \cong l$

$\\ \sf \orange{\leadsto l=336.5\ m}\ \; \bigstar$

Answer 2

$\bf{\gray{\underbrace{\blue{GIVEN:-}}}}$

• A wire has,

1. $\bf\red{Diameter}$ = 0.7 mm = 0.0007 m
2. $\bf\red{Resistivity}$ = 1.6 × $\bf{10^{-8}}$ Ω.m

$\bf{\gray{\underbrace{\blue{TO\:FIND:-}}}}$

• The length of the wire to make it's Resistance "14 Ω" .

$\bf{\gray{\underbrace{\blue{SOLUTION:-}}}}$

We have know that,

$\orange\bigstar\:\bf{\gray{\overbrace{\underbrace{\purple{R\:=\:\rho\:\dfrac{l}{A}\:}}}}}$----(1)

Where,

• $\bf\red{R}$ = Resistance

• $\bf\red{\rho}$ = Resistivity

• $\bf\red{l}$ = Length of the wire

• $\bf\red{A}$ = Area of the wire

According to the question,

• R = 14 Ω

• $\bf{\rho}$ = 1.6 × $\bf{10^{-8}}$ Ω.m

Now, to find the area of the wire we can use the following formula;

$\green\bigstar\:\bf{\gray{\overbrace{\underbrace{\purple{Area\:=\:\pi\:r^2\:}}}}}$

Where,

• $\bf{\pi}$ = 3.14

• r = radius = diameter/2 = 0.0007/2 = $\bf\pink{3.5\times{10^{-4}}\:m}$

$\rm{:\implies\:Area(A)\:=\:3.14\times{(3.5\times{10^{-4}})^2}\:}$

$\bf\blue{:\implies\:Area(A)\:=\:38.465\times{10^{-8}}\:m^2}$

Now, putting all these values in the equation (1);

$\bf\orange{:\implies\:l\:=\:\dfrac{RA}{\rho}\:}$

$\rm{:\implies\:l\:=\:\dfrac{14\times{38.465\times{10^{-8}}}}{1.6\times{10^{-8}}}\:}$

$\rm{:\implies\:l\:=\:\dfrac{538.51\times{10^{-8}}}{1.6\times{10^{-8}}}\:}$

$\bf\green{:\implies\:l\:=\:336.56\:m}$

$\red\therefore$ The length of the wire to make it's Resistance "14 Ω" is $\bf\red{336.56\:m}$ .