Resistances of a wire at \( 20^\circ\mathrm{C} \) and at \( 500^\circ\mathrm{C} \) are \( 20\ \Omega \) and \( 60\ \Omega \), respectively. The temperature at which its resistance becomes \( 25\ \Omega \) will be:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( 80^\circ\mathrm{C} \)
Over this range the resistance of the wire rises linearly with temperature. Taking 20 °C as the reference temperature:
\( R_{T} = R_{20}\,[1 + \alpha(T - 20^\circ\mathrm{C})] \)
From the two given values:
\( 60\ \Omega = 20\ \Omega\,[1 + \alpha \times 480^\circ\mathrm{C}] \)
\( \alpha \times 480^\circ\mathrm{C} = 2 \)
\( \displaystyle \alpha = \frac{2}{480}\ ^\circ\mathrm{C}^{-1} = \frac{1}{240}\ ^\circ\mathrm{C}^{-1} \)
For \( R_{T} = 25\ \Omega \):
\( \displaystyle 25 = 20\left[1 + \frac{T - 20}{240}\right] \quad\Rightarrow\quad \frac{T - 20}{240} = 0.25 \)
\( T - 20^\circ\mathrm{C} = 60^\circ\mathrm{C} \)
\( T = 80^\circ\mathrm{C} \)
Shortcut: With a linear rise, resistance and temperature change in direct proportion: 40 Ω for 480 °C means 5 Ω for 60 °C, so T = 20 °C + 60 °C = 80 °C.