Question

Plzz Guys Answer With full explanation _____ 人


Wire has resistance of 10Ω . A second wire of same material has length double and radius of cross section half that of wire . Find the resistance of second wire ___ ..​

Answer 1

$\large\underline{\bigstar \: \: {\sf Given-}}$

• Resistance of wire ${\sf R_1}$ = 10Ω
• Lenght of wire = $\ell$
• Radius of wire = r
• Area of Cross-section = ${\sf A_1}$

$\large\underline{\bigstar \: \: {\sf To \: Find -}}$

• Resistance of second wire.

$\large\underline{\bigstar \: \: {\sf Formula \: Used -}}$

$\implies\underline{\boxed{\sf R=\rho\dfrac{\ell}{A}}}$

R = Resistance

$\ell$ = Lenght of wire

A = Area of Cross-section

$\implies{\sf R\:\:\propto\:\:\dfrac{\ell}{A}}$

$\large\underline{\bigstar \: \: {\sf Solution-}}$

Area of wire = πr²

For second wire ,

Lenght of wire is doubled

Radius of wire is halfed

$\implies{\sf A_2 = π \times \left(\dfrac{r}{2}\right)^2}$

$\implies{\sf A_2=\dfrac{πr^2}{4}}$

____________________________

Resistance of first wire -

$\implies{\bf R_1=\dfrac{\ell}{πr^2} =10Ω\:\:\:\:\:(Given)}$

Resistance of second wire -

$\implies{\sf R_2=\dfrac{2\ell}{\dfrac{πr^2}{4}} }$

$\implies{\sf R_2=\dfrac{2\ell \times 4}{πr^2} }$

$\implies{\sf R_2=\dfrac{8 \times\ell}{πr^2} }$

$\implies{\sf R_2=8 \times R_1 }$

$\implies{\sf R_2=8 \times 10 }$

$\implies{\bf R_2=80Ω }$

$\large\underline{\bigstar \: \: {\sf Answer-}}$

Resistance of second wire is ${\bf 80Ω}$