Q 12-03-100JEE MainJEE Main 2024 (9 Apr, Shift 2)Easy
At room temperature ($27\,^\circ\text{C}$), the resistance of a heating element is $50\ \Omega$. The temperature coefficient of the material is $2.4\times10^{-4}\ ^\circ\text{C}^{-1}$. The temperature of the element, when its resistance is $62\ \Omega$, is ______ $^\circ\text{C}$.
Numerical value type. Enter your answer.
Answer: 1027
$$R = R_{27}\left[1 + \alpha(T - 27)\right] \;\Rightarrow\; 62 = 50\left[1 + 2.4\times10^{-4}(T - 27)\right]$$
$$T - 27 = \frac{0.24}{2.4\times10^{-4}} = 1000 \;\Rightarrow\; T = 1027\,^\circ\text{C}$$
Solution by Sreeraj P, M.Sc Physics