By increasing the temperature, resistivity of a conductor and a semiconductor
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → increases for a conductor and decreases for a semiconductor
Resistivity is \( \rho = m/(n e^{2} \tau) \): it depends on the number density of free charge carriers \( n \) and on the relaxation time \( \tau \) between collisions.
In a metal (conductor) \( n \) does not change with temperature, but the ions vibrate more, collisions become more frequent and \( \tau \) falls. So the resistivity increases.
In a semiconductor, heating breaks more covalent bonds and \( n \) rises very rapidly; this outweighs the fall in \( \tau \). So the resistivity decreases.
Option (1) decreases for a conductor and increases for a semiconductor — Incorrect. This is the reverse of the actual behaviour of both materials.
Option (2) increases for a conductor and decreases for a semiconductor — Correct. More collisions raise the resistivity of the metal; more charge carriers lower that of the semiconductor.
Option (3) decreases for both — Incorrect. The resistivity of a metal rises, not falls, with temperature.
Option (4) increases for both — Incorrect. The resistivity of a semiconductor falls with temperature (negative temperature coefficient).
NCERT (2026-27):
Temperature Dependence of Resistivity – “… Unlike metals, the resistivities of semiconductors decrease with increasing temperatures. … In a metal, \( n \) is not dependent on temperature to any appreciable extent and thus the decrease in the value of \( \tau \) with rise in temperature causes \( \rho \) to increase as we have observed.”