At temperature \( 25\ ^{\circ}\mathrm{C} \), the resistance of a heating element is \( 100\ \Omega \). What is the temperature of the element, if the resistance is found to be \( 108.5\ \Omega \)?
(Given: the temperature coefficient of resistance of the material of the resistor is \( 1.70\times10^{-4}\ ^{\circ}\mathrm{C^{-1}} \))
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( 525\ ^{\circ}\mathrm{C} \)
The resistance of a metallic element increases linearly with temperature, with the temperature coefficient of resistance \( \alpha \) as the fractional increase per degree.
\( \displaystyle R_{2} = R_{1}\left[1+\alpha(T_{2}-T_{1})\right] \)
\( \displaystyle T_{2}-T_{1} = \frac{R_{2}-R_{1}}{\alpha R_{1}} = \frac{108.5\,\Omega-100\,\Omega}{(1.70\times10^{-4}\,^{\circ}\mathrm{C^{-1}})(100\,\Omega)} \)
\( \displaystyle T_{2}-T_{1} = \frac{8.5}{1.70\times10^{-2}}\,^{\circ}\mathrm{C} = 500\,^{\circ}\mathrm{C} \)
\( T_{2} = 25\,^{\circ}\mathrm{C} + 500\,^{\circ}\mathrm{C} = 525\,^{\circ}\mathrm{C} \)
Note: NTA picked this question from an NCERT (2026-27) exercise and changed the figures. Do not ignore NCERT while practising numericals.
NCERT (2026-27):
Exercises – …
