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 ___ ..
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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 ___ ..
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[tex]\large\underline{\bigstar \: \: {\sf Given-}}[/tex]
[tex]\large\underline{\bigstar \: \: {\sf To \: Find -}}[/tex]
[tex]\large\underline{\bigstar \: \: {\sf Formula \: Used -}}[/tex]
[tex]\implies\underline{\boxed{\sf R=\rho\dfrac{\ell}{A}}}[/tex]
R = Resistance
[tex]\ell[/tex] = Lenght of wire
A = Area of Cross-section
[tex]\implies{\sf R\:\:\propto\:\:\dfrac{\ell}{A}}[/tex]
[tex]\large\underline{\bigstar \: \: {\sf Solution-}}[/tex]
Area of wire = πr²
For second wire ,
Lenght of wire is doubled
Radius of wire is halfed
[tex]\implies{\sf A_2 = π \times \left(\dfrac{r}{2}\right)^2}[/tex]
[tex]\implies{\sf A_2=\dfrac{πr^2}{4}}[/tex]
____________________________
Resistance of first wire -
[tex]\implies{\bf R_1=\dfrac{\ell}{πr^2} =10Ω\:\:\:\:\:(Given)}[/tex]
Resistance of second wire -
[tex]\implies{\sf R_2=\dfrac{2\ell}{\dfrac{πr^2}{4}} }[/tex]
[tex]\implies{\sf R_2=\dfrac{2\ell \times 4}{πr^2} }[/tex]
[tex]\implies{\sf R_2=\dfrac{8 \times\ell}{πr^2} }[/tex]
[tex]\implies{\sf R_2=8 \times R_1 }[/tex]
[tex]\implies{\sf R_2=8 \times 10 }[/tex]
[tex]\implies{\bf R_2=80Ω }[/tex]
[tex]\large\underline{\bigstar \: \: {\sf Answer-}}[/tex]
Resistance of second wire is [tex]{\bf 80Ω}[/tex]