JEE Main 2024 Physics question
Resistance of a wire at 0°C, 100°C and t°C is found to be 10 Ω, 10.2 Ω and 10.95 Ω respectively. The temperature t in Kelvin scale is .
Answer: 748
Solution
R = R₀(1 + αt): 10.2 = 10(1 + 100α) ⇒ α = 2 × 10⁻⁴; 10.95 = 10(1 + αt) ⇒ t = 475°C = 748 K.