To measure the temperature coefficient of resistivity $$\alpha$$ of a semiconductor, an electrical arrangement shown in the figure is prepared. The arm BC is made up of the semiconductor. The experiment is being conducted at 25°C and resistance of the semiconductor arm is $$3 \text{ m}\Omega$$. Arm BC is cooled at a constant rate of $$2 \text{ °C s}^{-1}$$. If the galvanometer G shows no deflection after 10 s, then $$\alpha$$ is

Treat it as a Wheatstone bridge. At balance:

$$\frac{AB}{BC}=\frac{AD}{DC}$$

Given:
AB = 0.8 mΩ
AD = 1 mΩ
DC = 3 mΩ

So,

Initially (at 25°C), BC = 3 mΩ
After cooling, it becomes 2.4 mΩ

step 2: temperature change

Cooling rate = 2 °C/s
Time = 10 s

$$\Delta T=20^{\circ}C$$

Final temperature = 25 − 20 = 5°C

step 3: use resistance-temperature relation

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

Here ΔT = −20°C:

$$2.4=3(1+\alpha(-20))$$

$$2.4=3(1-20\alpha)$$

$$0.8=1-20\alpha$$