At room temperature (27ºC), the resistance of a heating element is 50 Ω. The temperature coefficient of the material is 2.4 × 10⁻⁴ ºC⁻¹. The temperature of the element, when its resistance is 62 Ω, i s ______ ºC.
Physics · Current electricity · electric current ,current density and resistance
JEE Mains 2024 09 Apr Evening Shift

Question

At room temperature (27ºC), the resistance of a heating element is 50 Ω. The temperature coefficient of the material is 2.4 × 10⁻⁴ ºC⁻¹. The temperature of the element, when its resistance is 62 Ω, i s ______ ºC.

Physics · Current electricity · electric current ,current density and resistance

JEE Mains 2024 09 Apr Evening Shift

G1 Physics · Accepted Answer

Resistance-temperature relation: R = R₀[1 + αΔT] Given: R = 62 Ω R₀ = 50 Ω α = 2.4 × 10⁻⁴ °C⁻¹ Therefore: 62 = 50[1 + 2.4 × 10⁻⁴ ΔT] 62/50 - 1 = 2.4 × 10⁻⁴ ΔT 0.24 = 2.4 × 10⁻⁴ ΔT ΔT = 1000°C Initial temperature is 27°C. Therefore: T(final) = 27 + 1000 = 1027°C