QJEEM24Phf5ff: JEE Main 2024 Physics - Class 12 JEEM24Phf5ff Solution: At room temperature ( 27 ∘ C ) , the resistance of a heating…

We can start solving this problem by first understanding that the resistance of a material changes with temperature, and this change can be quantified using the temperature coefficient of resistance $\alpha$ . The relationship between the resistance of a material at any temperature $T$ and its resistance at a reference temperature $T_0$ is given by the formula:

$R = R_0(1 + \alpha(T - T_0))$

Where:

• $R$ is the resistance at temperature $T$ (in this case, $62 \Omega$ ).
• $R_0$ is the resistance at reference temperature $T_0$ (in this case, $50 \Omega$ ).
• $\alpha$ is the temperature coefficient of resistance (in this case, $2.4 \times 10^{-4} \, ^\circ\mathrm{C}^{-1}$).
• $T$ is the unknown temperature we need to find.
• $T_0$ is the reference temperature, given as $27^\circ \mathrm{C}$ .

By substituting the given values into the formula, we get:

$62 = 50(1 + 2.4 \times 10^{-4}(T - 27))$

First, divide both sides of the equation by 50:

$\frac{62}{50} = 1 + 2.4 \times 10^{-4}(T - 27)$

Then solve for $T$ :

$1.24 = 1 + 2.4 \times 10^{-4}(T - 27)$

$0.24 = 2.4 \times 10^{-4}(T - 27)$

$\frac{0.24}{2.4 \times 10^{-4}} = T - 27$

$1000 = T - 27$

$T = 1027 ^\circ\mathrm{C}$

Thus, the temperature of the element when its resistance is $62 \Omega$ is $1027^\circ\mathrm{C}$ .