A metal wire of resistance 3 Ω is elongated to make a uniform wire of double its previous length. This new…

1. $\frac{5}{3}\mathrm{\Omega }$
2. $\frac{12}{5}\mathrm{\Omega }$
3. $\frac{7}{2}\mathrm{\Omega }$
4. $\frac{5}{2}\mathrm{\Omega }$

Solution

We know, $R=\frac{\rho l}{A}=\frac{\rho l^2}{Al}=\frac{\rho l^2}{\mathrm{V}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{m}\mathrm{e}}$
$R\propto l^2$ , if the volume is constant.
$\therefore \mathrm{ }$ New resistance $=4R$

$R_1=\frac{4R}{360}\times 300=\frac{10R}{3}$
$R_2=\frac{4R}{360}\times 60=\frac{2R}{3}$
$\therefore$ Equivalent resistance $R_{\mathrm{e}\mathrm{q}}=\frac{R_1{R}_2}{R_1+R_2}$
$=\frac{\frac{10R}{3}\times \frac{2R}{3}}{\frac{10R}{3}+\frac{2R}{3}}=\frac{5R}{9}$
$=\frac{5}{3}\mathrm{ }\mathrm{\Omega }$

Asked in: JEE Main 2019 (09 Apr Shift 2)

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