Q128 / 180 · COMEDK 2025 (10 May Shift 3) · Physics

The resistance of a heating element is found to be $120 \Omega$ at room temperature which is $20^{\circ} \mathrm{C}$. If the temperature coefficient of the material of the resistor is $1.6 \times 10^{-4} C^{-1}$ and the resistance is found to be $160 \Omega$, the temperature of the element is:

  • A
    $1203^{\circ} \mathrm{C}$
  • Correct answer
    B
    $2103^{\circ} \mathrm{C}$
  • C
    $2310^{\circ} \mathrm{C}$
  • D
    $2013^{\circ} \mathrm{C}$

Solution

The resistance of a conductor at a temperature $T$ is given by the formula $R_T = R_0 [1 + \alpha(T - T_0)]$, where $R_0$ is the resistance at room temperature $T_0$, $R_T$ is the resistance at temperature $T$, and $\alpha$ is the temperature coefficient of resistance.

Given values are $R_0 = 120 \Omega$, $T_0 = 20^{\circ} \mathrm{C}$, $R_T = 160 \Omega$, and $\alpha = 1.6 \times 10^{-4} \mathrm{C}^{-1}$.

Substituting these values into the equation:

$160 = 120 [1 + 1.6 \times 10^{-4} (T - 20)]$

Dividing both sides by $120$:

$\dfrac{160}{120} = 1 + 1.6 \times 10^{-4} (T - 20)$

$\dfrac{4}{3} = 1 + 1.6 \times 10^{-4} (T - 20)$

$\dfrac{4}{3} - 1 = 1.6 \times 10^{-4} (T - 20)$

$\dfrac{1}{3} = 1.6 \times 10^{-4} (T - 20)$

$T - 20 = \dfrac{1}{3 \times 1.6 \times 10^{-4}}$

$T - 20 = \dfrac{1}{4.8 \times 10^{-4}} = \dfrac{10000}{4.8} \approx 2083.33$

$T = 2083.33 + 20 = 2103.33 \approx 2103^{\circ} \mathrm{C}$.

Answer: $2103^{\circ} \mathrm{C}$

Hence, the correct answer is option: 'B'.

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