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 centre, the equivalent resistance between these two points will be

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

Answer 1

The correct option is C $\frac{5}{3}\Omega$

We know that when we double the length of the wire by elongation , its volume will remain the same
volume of wire $V=Al=$ constant
$A$= cross sectional area , $l$= length )
Now, resistance

$R=\frac{\rho l}{A}$

$=\frac{\left(\rho l^2\right)}{V}\Rightarrow R\alpha l^2$

Hence, doubling the length makes the resistance four times
So, if

$R_{\text{initial}}=3\Omega$

then
$R_{final}=4\left(3\Omega \right)=12\Omega$

Total resistance of circular wire $=R_{final}$

Resistance of arc subtending angle
$\theta$ at centre

$=\frac{\theta }{2\pi }\times R_{\text{final}}$

Total resistance of larger part of circle as shown in figure below
Resistance of larger part of circle :

$\frac{\text{Angle subtended by the larger arc}}{2\pi }\times R_{final}$

$\Rightarrow \frac{5\pi }{(3\times 2\pi )}\times R_{\text{final}}=10\Omega$

Resistance of smaller part of circle :
$\frac{\text{Angle subtended by the smaller arc}}{2\pi }\times R_{f}inal$

$\Rightarrow \frac{\pi }{(3\times 2\pi )}\times R_{\text{final}}=2\Omega$

So, the equivalent resistance between $A$ and $B$ is

$R_{AB}=\frac{R_1{R}_2}{R_1+R_2}=\frac{10\times 2}{10+2}=\frac{5}{3}\Omega$

(parallel combination )

Hence, option (b) is correct.