Question

The wire used in the arrangement shown in the figure has a resistance of $r\Omega$ per metre. The equivalent resistance between points A and B is:

• A. $\left(\frac{6}{11}\right)r$
• B. $\frac{2\pi r}{(\pi +1)}$
• C. $\frac{6\pi r}{(16+3\pi )}$
• D. $\frac{3\pi r}{(10+3\pi )}$

Answer 1

The correct option is C $\frac{6\pi r}{(16+3\pi )}$

$\text{Step 1: Drawing the equivalent circuit[Refer figure]}$

All three parts of wire between A and B can be considered in parallel, as both of their ends connected to same points A and B.

$\text{Step 2: Calculation of resistance of all parts}$

Radius of circle $\left(R\right)=1m$

Resistance of unit length of wire $=r\Omega$

So,

Resistance of part $1\left(R_1\right)=r\times L_1$

$=r\times \frac{\pi R}{2}$ $=\frac{\pi r}{2}\Omega$

Resistance of part $2\left(R_3\right)=r\times L_2$

$=r\times 2R=2r\Omega$

Resistance of part $3\left(R_2\right)=r\times L_3$

$=r\times \frac{3\pi R}{2}=\frac{3\pi r}{2}\Omega$

$\text{Step 3: Effective resistance between point A and B -}$

$R_1,R_2$ and $R_3$ are in parallel arrangement as shown in figure $\left(2\right)$

So, $\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}$

$=\frac{1}{\left(r\frac{\pi }{2}\right)}+\frac{1}{\left(\frac{3r\pi }{2}\right)}+\frac{1}{\left(2r\right)}$

On solving, We get:

$R_{eq}=\frac{6\pi r}{16+3\pi }\Omega$
Hence Option $C$ is correct.