JEE Physics 2024 Question with Solution

Given: Initial resistance $R_0 = 50 \, \Omega$, final resistance $R = 62 \, \Omega$, temperature coefficient $\alpha = 2.4 \times 10^{-4} \, ^\circ\text{C}^{-1}$, and initial temperature $T_0 = 27^\circ \text{C}$.

Find: The temperature of the element when its resistance becomes $62 \, \Omega$.

Use the relation between resistance and temperature:

$$R = R_0 (1 + \alpha \Delta T)$$

Substituting the given values:

$$62 = 50(1 + 2.4 \times 10^{-4} \Delta T)$$ $$\frac{62}{50} = 1 + 2.4 \times 10^{-4} \Delta T$$ $$1.24 - 1 = 2.4 \times 10^{-4} \Delta T$$ $$0.24 = 2.4 \times 10^{-4} \Delta T$$ $$\Delta T = \frac{0.24}{2.4 \times 10^{-4}} = 1000^\circ \text{C}$$

Now calculate the final temperature:

$$T = T_0 + \Delta T = 27 + 1000 = 1027^\circ \text{C}$$

Therefore, the temperature of the element is $1027^\circ \text{C}$. The correct option is B.

Given: $R_0 = 50 \, \Omega$, $R = 62 \, \Omega$, $\alpha = 2.4 \times 10^{-4} \, ^\circ\text{C}^{-1}$, and $T_0 = 27^\circ \text{C}$.

Find: Final temperature $T$.

Rearrange the formula directly:

$$\Delta T = \frac{R - R_0}{\alpha R_0}$$

Substitute the values:

$$\Delta T = \frac{62 - 50}{(2.4 \times 10^{-4}) \cdot 50}$$ $$\Delta T = \frac{12}{0.012} = 1000^\circ \text{C}$$

Then,

$$T = 27 + 1000 = 1027^\circ \text{C}$$

This shortcut works because the resistance change is linear with temperature for small temperature coefficients. Therefore, the correct option is B.