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Step-by-Step Solution
Step 1: Write Down the Known Data
• Resistance of the conductor at temperature $15^\circ \text{C}$, $R_1 = 16\,\Omega$.
• Resistance of the conductor at temperature $100^\circ \text{C}$, $R_2 = 20\,\Omega$.
• We need to find the temperature coefficient of resistance, $\alpha$.
Step 2: Use the Formula for Temperature Dependence of Resistance
For a conductor with resistance $R_0$ at $0^\circ \text{C}$ (an assumed reference), the resistance $R$ at temperature $T$ is given by:
$$
R = R_0 \bigl(1 + \alpha\,T \bigr).
$$
Thus, for $T_1 = 15^\circ \text{C}$ and $T_2 = 100^\circ \text{C}$, we can write:
$$
R_1 = R_0 \bigl[1 + \alpha \times 15 \bigr],
$$
$$
R_2 = R_0 \bigl[1 + \alpha \times 100 \bigr].
$$
Step 3: Form the Ratio and Solve for $\alpha$
Divide the two expressions to eliminate $R_0$:
$$
\frac{R_1}{R_2} = \frac{1 + \alpha \times 15}{1 + \alpha \times 100}.
$$
Substitute $R_1 = 16\,\Omega$ and $R_2 = 20\,\Omega$:
$$
\frac{16}{20} = \frac{1 + 15\,\alpha}{1 + 100\,\alpha}.
$$
Simplify the fraction on the left:
$$
\frac{16}{20} = 0.8.
$$
So we have:
$$
0.8 = \frac{1 + 15\,\alpha}{1 + 100\,\alpha}.
$$
Cross-multiply and solve for $\alpha$:
$$
0.8 \bigl( 1 + 100\,\alpha \bigr) = 1 + 15\,\alpha.
$$
$$
0.8 + 80\,\alpha = 1 + 15\,\alpha.
$$
$$
80\,\alpha - 15\,\alpha = 1 - 0.8.
$$
$$
65\,\alpha = 0.2.
$$
$$
\alpha = \frac{0.2}{65} = 0.003\,^\circ\text{C}^{-1}.
$$
Step 4: State the Final Answer
The temperature coefficient of resistance of the conductor is
$$
\alpha = 0.003\,^\circ \text{C}^{-1}.
$$
