JEE Main 2024 — Current Electricity Question with Solution

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 $1027^{\circ }\mathrm{C}$.