A metal wire of resistance $3\mathrm{\Omega }$ is elongated to make a uniform wire of double its previous length. This new wire is now bent and the ends joined to make a circle. If two points on this circle make an angle ${60}^{\mathrm{o}}$ at the center, the equivalent resistance between these two points will be:

• Correct answer
  A

  $\frac{5}{3}\mathrm{\Omega }$
• B

  $\frac{12}{5}\mathrm{\Omega }$
• C

  $\frac{7}{2}\mathrm{\Omega }$
• D

  $\frac{5}{2}\mathrm{\Omega }$

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 $R_{\mathrm{e}\mathrm{q}}=\frac{R_1{R}_2}{R_1+R_2}$
$=\frac{\frac{10R}{3}\times \frac{2R}{3}}{\frac{10R}{3}+\frac{2R}{3}}=\frac{5R}{9}$
$=\frac{5}{3}\mathrm{ }\mathrm{\Omega }$

Hence, the correct answer is option: 'A'.