Question

The temperature coefficient of resistance of a coil of wire of resistance \(4 \Omega\) at \(30^\circ\text{C}\) and \(6 \Omega\) at \(70^\circ\text{C}\) (in per \(^\circ\text{C}\)) is

• A. 0.04
• B. 0.06
• C. 0.004
• D. 0.006
• E. 0.02

Hint:

If you use the approximate formula \(\alpha = \frac{R_2 - R_1}{R_1(t_2 - t_1)}\), you might get a slightly different value (\(2/160 = 0.0125\)). In competitive exams, always use the ratio method with \(R_0\) for more accuracy, or check which option matches your approximate result.

Answer 1

Answer: (E)

Step 1: Understanding the Concept:
The resistance of a metal conductor increases linearly with temperature. The temperature coefficient of resistance (\(\alpha\)) represents the fractional change in resistance per degree temperature rise.

Step 2: Key Formula or Approach:
Resistance at temperature \(t\) is \(R_t = R_0(1 + \alpha t)\), where \(R_0\) is resistance at \(0^\circ\text{C}\).

Step 3: Detailed Explanation:
From the given data:
\(R_{30} = R_0(1 + 30\alpha) = 4 \dots (i)\)
\(R_{70} = R_0(1 + 70\alpha) = 6 \dots (ii)\)
Dividing (ii) by (i):
\[ \frac{6}{4} = \frac{1 + 70\alpha}{1 + 30\alpha} \]
\[ 1.5(1 + 30\alpha) = 1 + 70\alpha \]
\[ 1.5 + 45\alpha = 1 + 70\alpha \]
\[ 1.5 - 1 = 70\alpha - 45\alpha \]
\[ 0.5 = 25\alpha \implies \alpha = \frac{0.5}{25} = \frac{1}{50} = 0.02 \]

Step 4: Final Answer:
The temperature coefficient of resistance is 0.02.