JEE Main 2024 Physics question
At room temperature (27°C), the resistance of a heating element is 50Ω. The temperature coefficient of the material is 2.4 × 10⁻⁴ °C⁻¹. The temperature of the element, when its resistance is 62 Ω, is ............. °C.
Answer: 1027
Solution
R = R₀(1 + αΔT): 62 = 50[1 + 2.4 × 10⁻⁴ΔT], so ΔT = 1000°C and T = 27 + 1000 = 1027°C.