By increasing the temperature, the specific resistance of a conductor and a semiconductor
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → increases and decreases, respectively
The resistivity is \( \rho = m/(n e^{2}\tau) \), where \( n \) is the number density of charge carriers and \( \tau \) the relaxation time.
In a conductor (metal) \( n \) is almost fixed, while a rise in temperature makes the electrons collide more often, so \( \tau \) falls and the resistivity increases.
In a semiconductor a rise in temperature frees many more carriers, so \( n \) increases rapidly, and this outweighs the fall in \( \tau \). The resistivity decreases.
So the resistivity increases for the conductor and decreases for the semiconductor.
NCERT (2026-27):
Temperature Dependence of Resistivity – “… Unlike metals, the resistivities of semiconductors decrease with increasing temperatures.”
Temperature Dependence of Resistivity – “… 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.”