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Step-by-Step Solution
Step 1: Identify the known quantities
• Each wire (iron and copper-nickel alloy) has the same length (say $l$).
• Diameter of each wire, $d = 2\,\text{mm} = 2 \times 10^{-3}\,\text{m}$.
• Resistivities (in $\Omega \,\text{m}$) need to be converted from $\mu\Omega\,\text{cm}$:
$\rho_{\text{Fe}} = 12\,\mu\Omega \cdot \text{cm} = 12 \times 10^{-6} \,\Omega \cdot \text{cm} = 12 \times 10^{-8}\,\Omega \cdot \text{m}$
$\rho_{\text{CuNi}} = 51\,\mu\Omega \cdot \text{cm} = 51 \times 10^{-6} \,\Omega \cdot \text{cm} = 51 \times 10^{-8}\,\Omega \cdot \text{m}$
• The required equivalent resistance, $R_{\text{eq}} = 3\,\Omega$ when the two wires are connected in parallel.
Step 2: Express the resistance of each wire
The resistance $R$ of a wire of length $l$, cross-sectional area $A$, and resistivity $\rho$ is given by
$$
R = \frac{\rho\,l}{A}.
$$
Since the diameter is $2 \times 10^{-3}\,\text{m}$, the radius $r$ is $1 \times 10^{-3}\,\text{m}$. Hence, the cross-sectional area $A$ is
$$
A = \pi r^2 = \pi \left(1 \times 10^{-3}\,\text{m}\right)^2 = \pi \times 10^{-6}\,\text{m}^2.
$$
Therefore:
$$
R_{\text{Fe}} = \frac{\rho_{\text{Fe}}\,l}{\pi \times 10^{-6}}, \quad
R_{\text{CuNi}} = \frac{\rho_{\text{CuNi}}\,l}{\pi \times 10^{-6}}.
$$
Step 3: Use parallel resistance formula
For two resistors $R_{\text{Fe}}$ and $R_{\text{CuNi}}$ connected in parallel, the equivalent resistance $R_{\text{eq}}$ is:
$$
\frac{1}{R_{\text{eq}}} = \frac{1}{R_{\text{Fe}}} + \frac{1}{R_{\text{CuNi}}}.
$$
We are given $R_{\text{eq}} = 3\,\Omega$, so
$$
\frac{1}{3} = \frac{1}{\frac{\rho_{\text{Fe}}\,l}{\pi \times 10^{-6}}}
+ \frac{1}{\frac{\rho_{\text{CuNi}}\,l}{\pi \times 10^{-6}}}.
$$
Step 4: Simplify and solve for $l$
Factor out common terms:
$$
\frac{1}{3}
= \frac{\pi \times 10^{-6}}{\rho_{\text{Fe}}\,l}
+ \frac{\pi \times 10^{-6}}{\rho_{\text{CuNi}}\,l}
= \pi \times 10^{-6} \left(\frac{1}{\rho_{\text{Fe}}} + \frac{1}{\rho_{\text{CuNi}}}\right)\frac{1}{l}.
$$
Rearrange to find $l$:
$$
l = \pi \times 10^{-6} \left(\frac{1}{\rho_{\text{Fe}}} + \frac{1}{\rho_{\text{CuNi}}}\right) \times 3.
$$
Substituting $\rho_{\text{Fe}} = 12 \times 10^{-8}\,\Omega\cdot\text{m}$ and
$\rho_{\text{CuNi}} = 51 \times 10^{-8}\,\Omega\cdot\text{m}$ into the expression,
we obtain (via detailed arithmetic) $l \approx 97\,\text{m}$.
Step 5: Conclude the answer
The required length of each wire (iron and copper-nickel alloy), when connected in parallel to yield an equivalent resistance of $3\,\Omega$, is
97 m.
