At room temperature (27), the resistance of a heating element is 50 Ω. The temperature coefficient of the material is 2.4 × 10⁻⁴ ⁻¹. The temperature of the element, when its resistance is 62 Ω, is __________. Numerical answer type. Work it out, then open the answer.

Answer: 1027. R = R₂₇[1 + α (T - 27)] 62 = 50[1 + 2.4 × 10⁻⁴(T - 27)] ⇒ 2.4 × 10⁻⁴(T - 27) = 0.24 T - 27 = 1000 ⇒ T = 1027 Answer: 1027.

At room temperature (27), the resistance of a heating element is 50… | JEE Main 2024 (9 Apr, Shift 2)

Q6JEE Main 2024 · 9 APR · SHIFT 2Easy

Ohm's Law, Resistivity & Temperature

At room temperature $(27^\circ\text{C})$ , the resistance of a heating element is $50\ \Omega$ . The temperature coefficient of the material is $2.4 \times 10^{-4}\ {}^\circ\text{C}^{-1}$ . The temperature of the element, when its resistance is $62\ \Omega$ , is __________ ${}^\circ\text{C}$ .

Numerical answer type. Work it out, then open the answer.

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Answer: 1027

$R = R_{27}[1 + \alpha (T - 27)]$
$62 = 50[1 + 2.4 \times 10^{-4}(T - 27)] \Rightarrow 2.4 \times 10^{-4}(T - 27) = 0.24$
$T - 27 = 1000 \Rightarrow T = 1027^\circ\text{C}$
Answer: $1027\ ^\circ\text{C}$ .

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