What equal length of an iron wire and a copper-nickel alloy wire, each of 2 mm diameter connected parallel to give an equivalent resistance of 3 $$\Omega$$?
(Given resistivities of iron and copper-nickel alloy wire are 12 $$\mu\Omega$$ cm and 51 $$\mu\Omega$$ cm respectively)

We start by recalling the resistance formula for a uniform wire:

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

where $$\rho$$ is the resistivity, $$L$$ is the length and $$A$$ is the cross-sectional area.

The two wires (iron and copper-nickel alloy) have

diameter $$d = 2\ \text{mm}=0.2\ \text{cm}\;,$$

so the radius is

$$r=\frac{d}{2}=1\ \text{mm}=0.1\ \text{cm}.$$

The area of a circle is $$A=\pi r^{2},$$ hence

$$A=\pi(0.1\ \text{cm})^{2}=\pi(0.01)\ \text{cm}^{2}=0.01\pi\ \text{cm}^{2}.$$

Both wires are required to have the same unknown length $$L$$ and are connected in parallel. For two resistances $$R_{1}$$ and $$R_{2}$$ in parallel we use

$$\frac{1}{R_{\text{eq}}}=\frac{1}{R_{1}}+\frac{1}{R_{2}}.$$

The given resistivities are

$$\rho_{\text{Fe}} = 12\ \mu\Omega\text{ cm}=12\times10^{-6}\ \Omega\text{ cm},$$

$$\rho_{\text{Cu-Ni}} = 51\ \mu\Omega\text{ cm}=51\times10^{-6}\ \Omega\text{ cm}.$$

Therefore the individual resistances are

$$R_{\text{Fe}}=\rho_{\text{Fe}}\frac{L}{A},\qquad R_{\text{Cu-Ni}}=\rho_{\text{Cu-Ni}}\frac{L}{A}.$$

The equivalent resistance is given to be

$$R_{\text{eq}} = 3\ \Omega.$$

Substituting the expressions for the two resistances into the parallel-combination formula, we get

$$\frac{1}{3}=\frac{1}{\rho_{\text{Fe}}\dfrac{L}{A}}+\frac{1}{\rho_{\text{Cu-Ni}}\dfrac{L}{A}} =\frac{A}{L}\left(\frac{1}{\rho_{\text{Fe}}}+\frac{1}{\rho_{\text{Cu-Ni}}}\right).$$

Now we isolate $$L$$:

$$\frac{1}{3}=\frac{A}{L}\left(\frac{\rho_{\text{Fe}}+\rho_{\text{Cu-Ni}}}{\rho_{\text{Fe}}\rho_{\text{Cu-Ni}}}\right)$$

$$\Rightarrow\;L=3A\frac{\rho_{\text{Fe}}+\rho_{\text{Cu-Ni}}}{\rho_{\text{Fe}}\rho_{\text{Cu-Ni}}}.$$

We substitute the numerical values:

$$A = 0.01\pi\ \text{cm}^{2},\qquad \rho_{\text{Fe}} = 12\times10^{-6}\ \Omega\text{ cm},\qquad \rho_{\text{Cu-Ni}} = 51\times10^{-6}\ \Omega\text{ cm}.$$

First, add the resistivities:

$$\rho_{\text{Fe}}+\rho_{\text{Cu-Ni}}=(12+51)\times10^{-6}=63\times10^{-6}\ \Omega\text{ cm}.$$

Next, multiply them:

$$\rho_{\text{Fe}}\rho_{\text{Cu-Ni}}= (12\times10^{-6})(51\times10^{-6})=612\times10^{-12}\ \Omega^{2}\text{ cm}^{2}.$$

Insert everything into the length expression:

$$L = 3(0.01\pi)\frac{63\times10^{-6}}{612\times10^{-12}} =0.03\pi\cdot63\times10^{-6}\times\frac{1}{612\times10^{-12}}.$$

Simplifying the powers of ten:

$$\frac{10^{-6}}{10^{-12}} = 10^{6},$$

so

$$L = 0.03\pi\cdot63\cdot10^{6}\times\frac{1}{612}.$$

Compute the numerical factor:

$$0.03\times63 = 1.89,$$

$$\frac{1.89}{612}=0.0030882,$$

$$0.0030882\times\pi\approx0.009702.$$

Therefore

$$L\approx0.009702\times10^{6}\ \text{cm}=9702\ \text{cm}.$$

Finally convert centimetres to metres:

$$L=\frac{9702}{100}\ \text{m}\approx97.0\ \text{m}.$$

Hence, the correct answer is Option A.