Question

A metal wire of resistance $3\Omega$ 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 $60^{\circ }$ at the center, the equivalent resistance between these two points will be:

• A. $\frac{5}{3}\Omega$
• B. $\frac{12}{5}\Omega$
• C. $\frac{7}{2}\Omega$
• D. $\frac{5}{2}\Omega$

Answer 1

Answer: (A)

We know, $R=\frac{\rho l}{A}=\frac{\rho l^2}{Al}=\frac{\rho l^2}{\text{Volume}}$
$R\propto l^2$ , if the volume is constant.
$\therefore$ 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 $R_{eq}=\frac{R_1{R}_2}{R_1+R_2}$
$=\frac{\frac{10R}{3}\times \frac{2R}{3}}{\frac{10R}{3}+\frac{2R}{3}}=\frac{5R}{9}$
$=\frac{5}{3}\Omega$