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GUJCET · Physics · Current Electricity
At room temperature \(\left(27^{\circ} C \right)\) the resistance of a heating element is \(100 \Omega\). What is the temperature of the element if the resistance is found to be \(137 \Omega\), given that the temperature coefficient of the material of the resistor is \(1.35 \times 10^{-40} C ^{-1}\).
- A \(2767^{\circ} C\)
- B \(1227^{\circ} C\)
- C \(1027^{\circ} C\)
- D \(2327^{\circ} C\)
Answer & Solution
Correct Answer
(A) \(2767^{\circ} C\)
Step-by-step Solution
Detailed explanation
(A)
\(\begin{array}{l}
R_\theta=R_0\left[1+\alpha\left(\theta-\theta_0\right)\right] \\
137=100[1+1.35 \left.\times 10^{-4}(\theta-27)\right] \\
\therefore 1.37=1+1.35 \times 10^{-4}(\theta-27) \\
\therefore 0.37=1.35 \times 10^{-4}(\theta-27) \\
\therefore \frac{0.37}{1.35 \times 10^{-4}}=\theta-27 \\
\therefore 2740.74=\theta-27 \\
\therefore \theta \approx 2767^{\circ} C
\end{array}\)
\(\begin{array}{l}
R_\theta=R_0\left[1+\alpha\left(\theta-\theta_0\right)\right] \\
137=100[1+1.35 \left.\times 10^{-4}(\theta-27)\right] \\
\therefore 1.37=1+1.35 \times 10^{-4}(\theta-27) \\
\therefore 0.37=1.35 \times 10^{-4}(\theta-27) \\
\therefore \frac{0.37}{1.35 \times 10^{-4}}=\theta-27 \\
\therefore 2740.74=\theta-27 \\
\therefore \theta \approx 2767^{\circ} C
\end{array}\)
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