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, $R=\frac{\rho l}{A}=\frac{\rho l^{2}}{A l}=\frac{\rho l^{2}}{\text{Volume}}$
$R \propto l^{2}$ , if the volume is constant.
$\therefore \, $ New resistance $=4R$

$R_{1}=\frac{4 R}{360}\times 300=\frac{10 R}{3}$
$R_{2}=\frac{4 R}{360}\times 60=\frac{2 R}{3}$
$\therefore $ Equivalent resistance $R_{e q}=\frac{R_{1} R_{2}}{R_{1} + R_{2}}$
$=\frac{\frac{10 R}{3} \times \frac{2 R}{3}}{\frac{10 R}{3} + \frac{2 R}{3}}=\frac{5 R}{9}$
$=\frac{5}{3} \, \Omega$